Sample Angle Puzzle - Free Printable

Let’s solve this step by step. We’re given a diagram with intersecting lines and some angle measures, plus one key fact: triangle BDC is isosceles because BC = DC (that means two sides are equal, so the base angles are equal).

We’re told:
- ∠BDC = 70°
- BC = DC → so triangle BDC is isosceles with vertex D
- ∠BGH = 40°

We need to find many other angles. Let’s go one by one, using geometry rules like:

→ Angles on a straight line add to 180°
→ Vertical angles are equal
→ In an isosceles triangle, base angles are equal
→ Triangle interior angles add to 180°
→ Corresponding or alternate angles if lines are parallel? Wait — we don’t know if any lines are parallel yet. But maybe we can figure out some relationships.

First, let’s focus on triangle BDC.

Since BC = DC, then angles at B and C in triangle BDC are equal. That is:

∠DBC = ∠BCD

And we know ∠BDC = 70°, so:

∠DBC + BCD + 70° = 180°
→ 2 × ∠DBC = 110°
→ ∠DBC = 55°
→ So ∠BCD = 55° too

So now we have:
∠DBC = 55°
∠BCD = 55°
∠BDC = 70°

Now look at point B. Line AE crosses line CI at point B. So angles around point B should follow vertical angle and linear pair rules.

At point B, we have angle ABC and angle DBC forming a straight line along CI? Wait — actually, looking at the diagram description (even though I can’t see it, based on standard labeling), points A-B-D-E are on one line? Or maybe not. Actually, from the labels:

Points:
Line AE goes through A, B, D, E? Probably not — more likely, AE is one line, and CJ is another, crossing at D? Hmm.

Actually, let’s reconstruct based on common puzzles like this.

Typically in such diagrams:

- There’s a big “X” shape made by two lines crossing: say, line AE and line CJ cross at D.
- Then there’s another line FG crossing them, hitting CI at G and CJ at H.
- And triangle BDC is formed where B is on AE and CI, D is intersection of AE and CJ, C is top point.

But since we’re told BC = DC, and ∠BDC=70°, we already used that.

Also, ∠BGH = 40°. Point G is on line CI and also on line FE (since F-G-H-E seems to be a line). So line FE crosses CI at G, and CJ at H.

So ∠BGH is the angle at G between points B, G, H. Since B is above G on line CI, and H is to the right on line FE, then ∠BGH is the angle inside triangle BGH or just at point G.

Actually, ∠BGH = 40° — that’s the angle at G between segments GB and GH.

GB is part of line CI going up to B, and GH is part of line FE going to H.

So at point G, we have line CI and line FE intersecting. So vertical angles and linear pairs apply.

Let me try to list all required angles and compute them logically.

---

Start with what we know for sure:

From triangle BDC (isosceles, BC=DC, ∠BDC=70°):

→ ∠DBC = 55°
→ ∠BCD = 55°

Now, ∠ABC is adjacent to ∠DBC along line CI? If points I-G-B-C are colinear (which they seem to be, since it's labeled as line CI with points G and B on it), then ∠ABC and DBC form a straight line? Not necessarily — depends on position.

Wait — point A is on the left, connected to B, which is on line CI. So AB is a segment from A to B, and BC is from B to C. So angle ABC is the angle at B between points A, B, C.

Similarly, angle DBC is at B between D, B, C.

If points A, B, D are colinear (on line AE), then angles ABC and DBC are adjacent and together make angle ABD? No — better to think:

Assume line AE passes through A, B, D, E — so A-B-D-E are colinear.

Then at point B, we have line AE (horizontal-ish) and line CI (going up to C and down to I).

So angle between AB and BC is ∠ABC.

Angle between DB and BC is ∠DBC = 55°.

Since A-B-D are colinear, then ∠ABC and DBC are supplementary? Only if C is on one side.

Actually, if you stand at point B, looking toward A (left), then turning to C (up-right), that’s ∠ABC.

Then from C to D (right along the line), that would be continuing past B to D.

Wait — perhaps ∠ABC and CBD are adjacent angles making up the straight angle along line AD? No.

Better approach: use vertical angles and linear pairs.

Let’s define:

At point B:

Lines intersecting: line AE (A-B-D-E) and line CI (C-B-G-I)

So four angles at B:

- Between A and C: ∠ABC
- Between C and D: ∠CBD = 55° (we found this)
- Between D and G: ∠DBG
- Between G and A: ∠GBA

Since A-B-D is straight, and C-B-G is straight, then:

∠ABC + ∠CBD = 180°? No — only if C is on the opposite side.

Actually, if you draw it mentally:

Point B has two lines crossing: horizontal line AE (left to right: A-B-D-E), and diagonal line CI (top to bottom: C-B-G-I).

So the four angles at B are:

Top-left: between A and C → ∠ABC
Top-right: between C and D → ∠CBD = 55°
Bottom-right: between D and G → ∠DBG
Bottom-left: between G and A → ∠GBA

Since A-B-D is straight, the sum of angles on one side is 180°.

Specifically, ∠ABC + CBD = angle from A to D via C — but that’s not straight unless C is on the line, which it’s not.

Actually, the straight line is A-B-D, so any ray from B not on that line will create two angles that add to 180° with the adjacent ones.

More clearly: the angle between BA and BD is 180° (straight line).

Ray BC is coming out from B, so it splits the plane.

The angle between BA and BC is ∠ABC.

The angle between BC and BD is ∠CBD = 55°.

Therefore, ∠ABC + CBD = 180°, because BA and BD are opposite rays.

Is that correct?

Yes! Because A-B-D is a straight line, so vector BA and vector BD are 180° apart.

Ray BC is somewhere in between, so ∠ABC (between BA and BC) plus ∠CBD (between BC and BD) equals 180°.

Therefore:

∠ABC + 55° = 180°
→ ∠ABC = 125°

Great!

Similarly, vertically opposite angles: at point B, the angle opposite to ∠ABC is ∠DBG (between D and G).

Because when two lines intersect, vertical angles are equal.

Lines AE and CI intersect at B.

So:

∠ABC and DBG are vertical angles → equal
∠CBD and ∠ABG are vertical angles → equal

We have ∠CBD = 55°, so ∠ABG = 55°

And ∠ABC = 125°, so ∠DBG = 125°

Wait — let me confirm:

When two lines intersect, they form two pairs of vertical angles.

Line 1: A-B-D (horizontal)
Line 2: C-B-G (diagonal)

Angles:

- Top: between A and C → ∠ABC
- Right: between C and D → ∠CBD
- Bottom: between D and G → ∠DBG
- Left: between G and A → ∠GBA

Vertical angles:

∠ABC and DBG are vertical → equal
∠CBD and ∠GBA are vertical → equal

Yes.

So:

∠ABC = ∠DBG = 125°
∠CBD = ∠GBA = 55°

Perfect.

So now we have:

∠ABC = 125°
∠DBG = 125°
∠ABG = 55° (same as ∠GBA)
∠DBC = 55° (given earlier)

Now, next: ∠ADC

Point D is intersection of lines AE and CJ.

We know ∠BDC = 70°, which is part of triangle BDC.

At point D, lines AE and CJ intersect.

So similar to point B, we can find angles.

First, in triangle BDC, we have points B, D, C.

∠BDC = 70°, which is the angle at D between B and C.

Now, line AE goes through A-B-D-E, so at D, the line continues to E.

Line CJ goes through C-D-J.

So at point D, the two lines are AE and CJ, intersecting.

Angle between B and C is ∠BDC = 70°.

Since A-B-D-E is straight, then angle between A and C at D would be adjacent.

Specifically, ∠ADC is the angle at D between A and C.

Since A-D-E is straight, and C is above, then ∠ADC and ∠CDE are adjacent angles adding to 180°? Not exactly.

Actually, ray DA and ray DE are opposite.

Ray DC is coming out.

So angle between DA and DC is ∠ADC.

Angle between DC and DE is ∠CDE.

And ∠ADC + ∠CDE = 180°, because DA and DE are straight line.

But we also know that ∠BDC = 70°, and since B is on DA (because A-B-D, so B is between A and D), then ray DB is same as ray DA? No — if A-B-D, then from D, ray DB is towards B, which is towards A, so ray DB is same direction as ray DA.

Actually, if points are A-B-D-E colinear in that order, then from D, the ray towards B is the same as ray towards A.

So ∠BDC is the angle between DB and DC, which is same as angle between DA and DC, which is ∠ADC.

Is that right?

If D is after B, and A-B-D, then from D, looking back, B is between A and D, so direction from D to B is same as D to A.

Yes — so ray DB is identical to ray DA.

Therefore, ∠BDC = ∠ADC = 70°

Is that correct?

∠BDC is angle at D in triangle BDC, between points B, D, C.

Since B and A are on the same ray from D (because A-B-D, so from D, going through B gets to A), yes, so the direction to B is same as to A.

Thus, ∠ADC = ∠BDC = 70°

Okay, so ∠ADC = 70°

Similarly, vertically opposite angle at D: between E and J, etc.

But let’s continue.

Next, ∠DBC we already have: 55°

∠BCD = 55° (from isosceles triangle)

∠EDC: this is angle at D between E, D, C.

Since A-D-E is straight, and ∠ADC = 70°, then ∠EDC = 180° - 70° = 110°

Because they are adjacent angles on a straight line.

So ∠EDC = 110°

Now, moving to point G.

We’re given ∠BGH = 40°

Point G is on line CI and on line FE (F-G-H-E)

So at G, lines CI and FE intersect.

∠BGH is the angle at G between B, G, H.

Since B is on CI above G, and H is on FE to the right, so ∠BGH is the angle in the "northeast" quadrant at G.

Since lines CI and FE intersect at G, we can find other angles.

First, ∠BGH = 40°

This is the angle between ray GB and ray GH.

Ray GB is upward along CI, ray GH is rightward along FE.

Then, the vertical angle to ∠BGH would be the angle between ray GI and ray GF (down-left), which is ∠IGF or something.

But let’s label properly.

At point G:

Lines: CI (C-G-I) and FE (F-G-H-E)

So four angles:

- Between C and F: ∠CGF
- Between F and I: ∠FGI
- Between I and H: ∠IGH
- Between H and C: ∠HGC

Standard: when two lines intersect, vertical angles are equal.

∠BGH is same as ∠CGH? Point B is on CG, since C-B-G-I, so from G, ray GB is same as ray GC? No.

If points are C-B-G-I colinear, with B between C and G, then from G, ray GB is towards B, which is towards C, so ray GB is same as ray GC.

Yes — so ray GB = ray GC.

Therefore, ∠BGH is the angle between GC and GH, which is ∠CGH.

So ∠CGH = 40°

Then, since lines CI and FE intersect at G, the vertical angle to ∠CGH is ∠IGF (between I and F).

So ∠IGF = 40°

Also, adjacent angles: ∠CGH and ∠HGI are adjacent on line CI? No.

On line CI, from C to I, so at G, the straight line is C-G-I.

Ray GH is coming out.

So angle between GC and GH is 40°, then angle between GH and GI should be 180° - 40° = 140°, because GC and GI are opposite rays.

Yes!

So ∠HGI = 180° - ∠CGH = 180° - 40° = 140°

Similarly, ∠FGI: since F-G-H is straight, ray GF and GH are opposite.

So angle between GI and GF: since ∠HGI = 140°, and GF is opposite to GH, then ∠FGI = 180° - 140° = 40°? Let's think.

At point G:

- Ray GC and GI are opposite (line CI)
- Ray GF and GH are opposite (line FE)

Angle between GC and GH is 40° (∠CGH)

Then angle between GH and GI: since GC to GI is 180°, and GC to GH is 40°, so GH to GI is 140° → ∠HGI = 140°

Angle between GI and GF: since GH to GF is 180° (straight line), and GH to GI is 140°, so GI to GF is 40° → ∠IGF = 40°

Which matches vertical angle.

Angle between GF and GC: since GF to GH is 180°, GH to GC is 40° (but in other direction), actually, from GF to GC: going from GF to GH is 180°, then GH to GC is 40°, but that's not direct.

Better: the four angles at G:

- ∠CGF: between C and F
- ∠FGI: between F and I
- ∠IGH: between I and H
- ∠HGC: between H and C

We have ∠HGC = 40° (same as ∠BGH)

Then ∠FGI = vertical to ∠HGC = 40°

∠CGF = adjacent to ∠HGC on line FE? On line FE, from F to H, so at G, ray GF and GH opposite.

So angle between GC and GF: since GC to GH is 40°, and GH to GF is 180°, so GC to GF is 40° + 180° = 220°? No, angles around point are 360°.

Easier: the angle between GC and GF is the supplement if they are on a straight line, but they're not.

From the intersection:

The two pairs of vertical angles:

Pair 1: ∠CGH and ∠IGF = 40° each

Pair 2: ∠CGF and ∠IGH

And since total around point is 360°, and 40+40=80, so the other two sum to 280°, so each is 140° if equal, but are they?

In general, when two lines intersect, adjacent angles are supplementary.

So ∠CGH + ∠CGF = 180°? Are they adjacent?

Ray GC, then to GH and to GF.

Actually, rays from G: GC, GH, GI, GF.

Order around the point: depending on diagram, but typically, if CI is one line, FE is another, crossing.

Assume that from G, going clockwise: GC, then GH, then GI, then GF, back to GC.

Then angle between GC and GH is 40°

Between GH and GI: since GI is opposite to GC, and GH is 40° from GC, then from GH to GI is 180° - 40° = 140°? Let's calculate.

The angle from GC to GI is 180° (straight line).

If GH is 40° from GC, then from GH to GI is 180° - 40° = 140°.

Similarly, from GI to GF: since GF is opposite to GH, and GH to GI is 140°, then GI to GF is 180° - 140° = 40°? No.

From GI to GF: since GF is 180° from GH, and GI is 140° from GH (in the other direction), so angle between GI and GF is |180° - 140°| = 40°, but only if they are on the same side.

Actually, the angle between GI and GF is the same as between GC and GH, because vertical angles.

Earlier I said ∠IGF = 40°, which is angle at G between I, G, F, so that's between rays GI and GF.

Yes, and it's vertical to ∠CGH, so 40°.

Then the remaining angles: between GC and GF, and between GH and GI.

Angle between GC and GF: this is ∠CGF.

Since from GC to GH is 40°, and from GH to GF is 180° (because F-G-H straight), so from GC to GF is 40° + 180° = 220°, but that's reflex; the smaller angle is 360° - 220° = 140°.

Similarly, directly: since GC and GI are straight, and GF is 40° from GI (because ∠IGF=40°), so from GC to GF is 180° - 40° = 140°.

Yes.

So ∠CGF = 140°

Similarly, ∠IGH = 140° (vertical to ∠CGF)

So summary at G:

∠CGH = 40°
∠IGF = 40°
∠CGF = 140°
∠IGH = 140°

Now, the problem asks for ∠FGI, which is same as ∠IGF = 40°

And ∠FGH: but F-G-H is straight, so ∠FGH is 180°, but probably not asked.

Looking at the list:

We need ∠FGI = 40° (as above)

Also, ∠BDH, ∠DEH, etc.

Now, let's find ∠BDH.

Point D and H.

First, where is H? H is on CJ and on FE.

CJ is the line from C to J, passing through D and H.

FE is from F to E, passing through G and H.

So at H, lines CJ and FE intersect.

We need to find angles at H.

But first, do we know any angles at H?

We might need to use triangle or other relations.

Notice that we have points B, G, H.

We know ∠BGH = 40°, and we have points B, G, H forming triangle BGH? Possibly.

In triangle BGH, we know angle at G is 40°.

Do we know other angles?

At point B, we have angle ∠DBG = 125°, but that's between D, B, G.

In triangle BGH, vertices B, G, H.

At B, the angle would be between G, B, H.

But we have ray BD and BG.

From earlier, at B, we have ∠DBG = 125°, which is between D, B, G.

But H is not on that; H is on the other side.

Perhaps we can find angle at B in triangle BGH.

Ray BH: but we don't have direct info.

Another approach: perhaps lines are parallel? But not stated.

Maybe use the fact that we have transversal.

Let's list all required angles and see which we can find.

Required:

Column 1:
∠ABC = 125° (found)
∠ADC = 70° (found)
∠DBC = 55° (found)
∠BCD = 55° (found)
∠EDC = 110° (found)

Column 2:
∠ABG = 55° (found, vertical to ∠DBC)
∠BDH = ?
∠DEH = ?
∠DBG = 125° (found)
∠FGB = ?

Column 3:
∠GHD = ?
∠GHI = ?
∠HGI = 140° (found)
∠IGI = ? wait, ∠IGI doesn't make sense; probably typo, should be ∠IGF or something. Looking back:

In the original:

∠GHD = ______
∠GHI = ______
∠HGI = ______ (we have 140°)
∠IGI = ______ — this must be a typo. Probably ∠IGF or ∠FGI. In the list it's ∠IGI, but that doesn't make sense. Looking at the image description, it's likely ∠IGF or ∠FGI. In our case, we have ∠FGI = 40°.

Similarly, ∠FGI is listed separately.

In the user's input:

After ∠HGI = ______, it's ∠IGI = ______, but that must be a mistake. Probably it's ∠IGF or something. Looking at standard, likely it's ∠IGF, which is 40°.

But in the list, later there is ∠FGI, so perhaps ∠IGI is a typo for ∠IGF.

To avoid confusion, let's assume it's ∠IGF = 40°.

Similarly, ∠FGI is also listed, which is the same.

In the original text:

"∠GHD = ______
∠GHI = ______
∠HGI = ______
∠IGI = ______
∠FGI = ______"

∠IGI is probably a typo, and should be ∠IGF or ∠FGI. Since ∠FGI is listed separately, perhaps ∠IGI is meant to be ∠IGF, but it's redundant.

Perhaps it's ∠JGH or something. To resolve, let's look at the context.

Later in column 2, there is ∠FGB, which is at G between F, G, B.

We have that.

For now, let's proceed with what we have.

So far:

∠ABC = 125°
∠ADC = 70°
∠DBC = 55°
∠BCD = 55°
∠EDC = 110°
∠ABG = 55°
∠DBG = 125°
∠HGI = 140°
∠FGI = 40° (assuming ∠IGI is typo and same as ∠FGI or ∠IGF)

Now, need ∠BDH, ∠DEH, ∠FGB, ∠GHD, ∠GHI, etc.

Let's find ∠FGB.

At point G, ∠FGB is the angle between F, G, B.

Ray GF and ray GB.

From earlier, ray GB is same as ray GC.

Ray GF is to F.

We have ∠CGF = 140°, which is between C, G, F, so same as between B, G, F since B is on GC.

So ∠FGB = ∠CGF = 140°

Yes.

So ∠FGB = 140°

Now, similarly, ∠BDH.

Point D and H.

H is on CJ and FE.

CJ is the line from C to J, passing through D and H.

So points C-D-H-J are colinear? Probably, since it's a straight line CJ.

Similarly, FE is F-G-H-E.

So at H, lines CJ and FE intersect.

We need to find angles at H.

But we don't have direct measure, but perhaps we can use triangle or other.

Notice that we have points B, D, H.

Or perhaps consider triangle BDH or something.

Another idea: perhaps use the fact that we have transversal and corresponding angles, but no parallel lines given.

Maybe assume that some lines are parallel, but not stated.

Let's look at angle at H.

We know that at G, we have angles, and G and H are on the same line FE.

So on line FE, from F to E, passing through G and H.

So points F-G-H-E are colinear.

Therefore, at any point on this line, the straight line is 180°.

Now, at H, we have line CJ crossing it.

So similar to G, but we need a reference.

Perhaps we can find angle between CJ and FE at H.

But we don't have it yet.

Another approach: consider triangle BGH.

Points B, G, H.

We know angle at G: ∠BGH = 40°

Now, what is angle at B in this triangle?

At point B, the angle of triangle BGH is between points G, B, H.

From earlier, at B, we have several rays.

Ray BG is down to G (along CI).

Ray BH: but H is not on any known ray from B.

However, we can find the direction.

From B, we have ray BD (to D, along AE), ray BC (to C, up), ray BG (to G, down), ray BA (to A, left).

H is on FE, which is the line from F to E, passing through G and H.

From B, to H, it's a different ray.

Perhaps we can find the angle between BG and BH.

But we don't know.

Notice that line FE is straight, and we have points G and H on it.

At G, we have the angle between CG and GH is 40°.

CG is part of CI, which is the same line as CB extended.

So the line CI makes a 40° angle with FE at G.

Since FE is straight, and CI is straight, then at every point, the angle between them is the same, but only if they are not parallel, which they aren't.

Actually, the angle between two intersecting lines is constant, but here CI and FE intersect at G, so at H, which is on FE, but not on CI, so the angle at H between CJ and FE may be different if CJ is not the same as CI, but in this case, CJ is the same line as CI? No.

Line CI is from C to I, passing through B and G.

Line CJ is from C to J, passing through D and H.

Are CI and CJ the same line? Only if I, G, B, C, D, H, J are colinear, which they're not, because at C, it branches? No, in the diagram, likely CI and CJ are the same line? Let's think.

Points: C is common, then from C, one line goes to I passing through B and G, another line goes to J passing through D and H.

But in a typical such puzzle, often C is a vertex, and there are two lines from C: one to I, one to J, but in this case, since B and D are on different lines, probably CI and CJ are different lines.

But in the name, it's "line CI" and "line CJ", but in the diagram, it might be that C, B, G, I are on one line, and C, D, H, J are on another line.

Yes, that makes sense.

So line 1: C-B-G-I
Line 2: C-D-H-J
And they intersect at C.

Oh! I think I missed that.

In the diagram, lines CI and CJ both start from C, so they are two different lines from C.

So at point C, we have multiple lines.

Previously, we had triangle BDC, with B on one line, D on another.

So line CB is part of line CI, line CD is part of line CJ.

And they meet at C.

In triangle BDC, we have points B, D, C, with BC = DC, so isosceles.

At point C, the angle of the triangle is ∠BCD = 55°, which is the angle between CB and CD.

So the angle between line CI and line CJ at C is 55°.

Now, line FE intersects CI at G and CJ at H.

So FE is a transversal cutting the two lines CI and CJ.

At G, on CI, the angle between CI and FE is 40°, as given by ∠BGH = 40°, which is the angle between GB (which is along CI) and GH (along FE).

Since GB is along CI, and GH along FE, so the acute angle between the lines is 40°.

Similarly, at H, on CJ, the angle between CJ and FE can be found if we know the angle between CI and CJ.

The angle between CI and CJ at C is 55°.

FE is a line cutting both.

So the angle that FE makes with CI is 40° at G.

Then, since CI and CJ form 55° at C, the angle that FE makes with CJ at H can be calculated using the triangle or the directions.

Consider the triangle formed by C, G, H.

Points C, G, H.

G is on CI, H on CJ, and GH is part of FE.

So triangle CGH.

In this triangle, we know angle at G: between CG and GH.

CG is along CI, GH along FE, and we have ∠CGH = 40°.

Angle at C: between CG and CH.

CG is along CI, CH is along CJ, and angle between CI and CJ is ∠BCD = 55°, which is the angle at C in triangle BDC, but in triangle CGH, it's the same angle, since G is on CI, H on CJ, so angle at C in triangle CGH is the same as angle between CI and CJ, which is 55°.

Is that correct? Yes, because rays CG and CH are the same as rays CB and CD, since G is on CB extended, H on CD extended.

So in triangle CGH, angle at C is 55°, angle at G is 40°, so angle at H is 180° - 55° - 40° = 85°.

So ∠CHG = 85°

Now, ∠CHG is the angle at H in triangle CGH, between points C, H, G.

So between rays HC and HG.

HC is along CJ (since H is on CJ, and C is end, so ray HC is towards C, which is along CJ).

HG is along FE, towards G.

So at point H, the angle between CJ and FE is 85°.

Specifically, ∠CHG = 85°, which is the angle between CH and GH.

Now, since lines CJ and FE intersect at H, we can find other angles at H.

First, ∠GHD: this is angle at H between G, H, D.

D is on CJ, beyond H or before? Since C-D-H-J, probably D is between C and H, or H between D and J.

In the labeling, likely C-D-H-J, so from C to J: C, then D, then H, then J.

So at H, ray HD is towards D, which is towards C, so same as ray HC.

Ray HG is towards G.

So ∠GHD is the angle between GH and HD, which is the same as between GH and HC, since HD and HC are opposite? No.

If C-D-H-J, then from H, ray HD is towards D, which is towards C, so same direction as ray HC.

Yes, so ray HD = ray HC.

Therefore, ∠GHD = ∠GHC = 85°

Same as ∠CHG.

So ∠GHD = 85°

Now, vertically opposite or adjacent.

At H, lines CJ and FE intersect.

So angles:

- Between C and F: ∠CHF
- Between F and J: ∠FHJ
- Between J and G: ∠JHG
- Between G and C: ∠GHC = 85°

Since CJ is straight, and FE is straight.

∠GHC = 85°, which is between GH and HC.

Then, the adjacent angle on the straight line CJ: between HC and HJ is 180°, so angle between GH and HJ is 180° - 85° = 95°? Let's see.

From ray HC to ray HJ is 180° (straight line).

Ray HG is 85° from HC, so from HG to HJ is 180° - 85° = 95°.

So ∠GHJ = 95°

Similarly, vertically opposite to ∠GHC is ∠FHJ, so ∠FHJ = 85°

And ∠CHF = 95° (vertical to ∠GHJ)

Now, the problem asks for ∠GHI.

What is I? I is on line CI, which is different from CJ.

∠GHI is angle at H between G, H, I.

But I is not on the same line; it's on the other line.

So rays HG and HI.

HI is not defined; probably it's a typo or mislabel.

Looking back at the original: "∠GHI = ______"

But in the diagram, I is on the bottom left, H is on the right, so likely it's ∠GHJ or something.

Perhaps it's ∠GHD, but that's already listed.

Another possibility: in some notations, I might be used for a point, but here it's probably ∠GHJ or ∠FHG.

To resolve, let's assume that ∠GHI is meant to be ∠GHJ, which is 95°, as above.

Similarly, ∠HGI we have as 140°, but that's at G.

For ∠IGI, likely typo, skip or assume 40°.

Now, ∠BDH.

Point B, D, H.

B and D are on line AE, H is on CJ.

So angle at D between B, D, H.

At D, we have lines AE and CJ intersecting.

We know ∠BDC = 70°, which is between B, D, C.

C is on CJ, so ray DC is along CJ.

H is also on CJ, beyond D or before.

If C-D-H-J, then from D, ray DC is towards C, ray DH is towards H, which is opposite direction.

So ray DH is opposite to ray DC.

Therefore, angle between DB and DH.

DB is along AE towards B, which is towards A.

DH is along CJ towards H.

Since ray DC is towards C, and ray DH is opposite, so angle between DB and DH is 180° - angle between DB and DC.

Angle between DB and DC is ∠BDC = 70°.

So angle between DB and DH is 180° - 70° = 110°.

Because DC and DH are straight line.

So ∠BDH = 110°

Similarly, ∠DEH.

Point D, E, H.

E is on AE, beyond D, so ray DE is opposite to DB.

H is on CJ.

So at D, angle between DE and DH.

Ray DE is opposite to DB.

Ray DH is as above.

Angle between DB and DH is 110°, as above.

Since DE is opposite to DB, angle between DE and DH is 180° - 110° = 70°.

Because DB and DE are straight line.

So ∠DEH = 70°? But E and H are not necessarily connected, but the angle at D in triangle DEH or just the angle.

∠DEH is angle at E? No, notation ∠DEH means angle at E between D, E, H.

I think I confused.

In standard notation, ∠DEH means the angle at E formed by points D, E, H.

So vertex at E.

Similarly, ∠BDH is angle at D between B, D, H.

For ∠DEH, it's at E.

So let's clarify.

∠BDH: vertex D, rays DB and DH.

We have that as 110°, as above.

∠DEH: vertex E, rays ED and EH.

E is on line AE, and on line FE (since F-G-H-E).

So at E, lines AE and FE intersect.

We need angle between ED and EH.

ED is along AE towards D, which is towards A.

EH is along FE towards H, which is towards G and F.

So at E, the two lines are AE and FE.

We need the angle between them.

Do we know any angle at E?

Not directly, but perhaps from other parts.

Notice that we have points D, E, H.

In particular, we have triangle DEH or something.

From earlier, at D, we have angle between DE and DH is 70°, as I calculated for the direction, but that's at D, not at E.

For ∠DEH, it's at E.

So let's find the angle at E between lines AE and FE.

We know that at G, on FE, the angle with CI is 40°, but CI is not related directly.

Perhaps use the fact that we have the whole figure.

Another way: consider that line FE is straight, and we have points G and H on it, and we know angles at G and H with the other lines.

At G, angle between FE and CI is 40°.

At H, angle between FE and CJ is 85°, as we found.

But CI and CJ are different lines.

To find angle at E, we need more.

Perhaps assume that the lines are such that we can find.

Let's list what we have and see.

Perhaps for ∠DEH, it is the same as angle at E in the configuration.

Notice that points D, E, H form a triangle.

We know some angles.

At D, in triangle DEH, angle at D is between ED and HD.

As above, ray ED is along AE towards D, but from D, ray DE is towards E.

Earlier, at D, angle between DE and DH.

Ray DE is towards E, ray DH is towards H.

From earlier calculation, since ray DB is towards B (left), ray DE is towards E (right), so opposite.

Ray DH is towards H, which is along CJ away from C.

Angle between DB and DH is 110°, as calculated.

Since DE is opposite to DB, angle between DE and DH is 180° - 110° = 70°.

So in triangle DEH, angle at D is 70°.

Now, at H, angle between HD and HE.

HD is along CJ towards D, which is towards C.

HE is along FE towards E, which is towards G and F.

At H, we have angle between HC and HG is 85°, as in triangle CGH.

HC is towards C, HG is towards G.

HE is opposite to HG, since F-G-H-E, so from H, ray HE is towards E, opposite to ray HG (towards G).

So ray HE = opposite of ray HG.

Therefore, angle between HC and HE is 180° - angle between HC and HG = 180° - 85° = 95°.

Because HG and HE are straight line.

So in triangle DEH, at H, angle between HD and HE.

HD is same as HC (since towards C), so angle between HD and HE is 95°.

So angle at H in triangle DEH is 95°.

Then, angle at E is 180° - angle at D - angle at H = 180° - 70° - 95° = 15°.

So ∠DEH = 15°

Great.

Now, finally, ∠FGB we have as 140°.

Now, for the last few.

∠GHI: as discussed, likely ∠GHJ = 95°, as above.

∠HGI = 140° (at G)

∠IGI: probably typo, perhaps ∠IGF = 40°, or skip.

In the list, it's "∠IGI = ______", but that doesn't make sense. Perhaps it's ∠JGH or something. To match, maybe it's ∠FGI, but that's listed separately.

Looking at the original user input:

"∠GHD = ______
∠GHI = ______
∠HGI = ______
∠IGI = ______
∠FGI = ______"

And earlier "∠FGI = ______" is also in column 3.

Probably "∠IGI" is a typo and should be "∠IGF" or "∠FGI", but since "∠FGI" is listed, perhaps "∠IGI" is meant to be "∠JGH" or "∠FHG".

To resolve, let's assume that "∠IGI" is a mistake, and it should be "∠IGF" = 40°, but that's same as ∠FGI.

Perhaps it's "∠CGH" but that's 40°.

Another possibility: in some fonts, I and J look similar, so "∠IGI" might be "∠JGH" or "∠JGI".

But to be safe, let's calculate what we can.

We have ∠GHD = 85° (as above)

∠GHI: if I is on the other line, perhaps it's the angle between G, H, and I, but I is not connected.

Perhaps it's ∠GHD, but that's already there.

Another thought: in the diagram, point I is on the bottom, so from H to I might be a line, but not specified.

Perhaps "∠GHI" is meant to be "∠GHJ" = 95°, as we have.

Similarly, "∠IGI" might be "∠IGF" = 40°.

And "∠FGI" = 40°.

So let's set:

∠GHD = 85°
∠GHI = 95° (assuming ∠GHJ)
∠HGI = 140°
∠IGI = 40° (assuming ∠IGF)
∠FGI = 40°

But ∠IGI and ∠FGI are both 40°, which is fine.

Now, for completeness, let's list all.

Also, in column 2, we have ∠FGB = 140°, already have.

Now, let's compile the answers.

First column:
∠ABC = 125°
∠ADC = 70°
∠DBC = 55°
∠BCD = 55°
∠EDC = 110°

Second column:
∠ABG = 55°
∠BDH = 110°
∠DEH = 15°
∠DBG = 125°
∠FGB = 140°

Third column:
∠GHD = 85°
∠GHI = 95° (assumed)
∠HGI = 140°
∠IGI = 40° (assumed)
∠FGI = 40°

Now, to confirm, let's see if all make sense.

For ∠DEH = 15°, we calculated from triangle DEH: angles 70° at D, 95° at H, so 15° at E.

Similarly, others seem consistent.

So final answers:

Final Answer:
∠ABC = 125°
∠ADC = 70°
∠DBC = 55°
∠BCD = 55°
∠EDC = 110°
∠ABG = 55°
∠BDH = 110°
∠DEH = 15°
∠DBG = 125°
∠FGB = 140°
∠GHD = 85°
∠GHI = 95°
∠HGI = 140°
∠IGI = 40°
∠FGI = 40°