Solved Name: Geometry Worksheet 4 - 1 Date Period For each | Chegg.com - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Problem 1

Given: Parallelogram $ABCD$. Side $DC = 8$, Side $BC = 10$, Angle $B = 62^\circ$.

1. Find side $AB$: In a parallelogram, opposite sides are equal in length. Side $AB$ is opposite to side $DC$. Since $DC = 8$, then $AB = 8$.
2. Find side $AD$: Side $AD$ is opposite to side $BC$. Since $BC = 10$, then $AD = 10$.
3. Find angle $A$ ($m\angle A$): Consecutive angles (angles next to each other) in a parallelogram add up to $180^\circ$. So, $\angle A + \angle B = 180^\circ$.
$$180^\circ - 62^\circ = 118^\circ$$
So, $m\angle A = 118^\circ$.
4. Find angle $D$ ($m\angle D$): Opposite angles are equal. Angle $D$ is opposite to angle $B$. Since $\angle B = 62^\circ$, then $m\angle D = 62^\circ$. (Alternatively, $\angle D$ and $\angle A$ are consecutive, so $180^\circ - 118^\circ = 62^\circ$).

Problem 2

Given: Parallelogram $ABCD$ with an extension to point $E$. Angle $B = 44^\circ$.

1. Find angle $A$ ($m\angle A$): Consecutive angles add up to $180^\circ$.
$$180^\circ - 44^\circ = 136^\circ$$
So, $m\angle A = 136^\circ$.
2. Find angle $BCD$ ($m\angle BCD$): Opposite angles are equal. Angle $BCD$ (which is angle $C$) is opposite to angle $A$. Since we found $\angle A = 136^\circ$, then $m\angle BCD = 136^\circ$.
3. Find angle $CDE$ ($m\angle CDE$): This is an exterior angle. In a parallelogram, an exterior angle is equal to the interior opposite angle. The interior opposite angle to $\angle CDE$ is $\angle B$. Since $\angle B = 44^\circ$, then $m\angle CDE = 44^\circ$.
*(Another way: Angle $ADC$ is opposite to Angle $B$, so $\angle ADC = 44^\circ$. Angles on a straight line add to $180^\circ$, so $\angle CDE = 180^\circ - 44^\circ = 136^\circ$? Wait, let's look at the diagram carefully. Angle $B$ and Angle $ADC$ are opposite. So $\angle ADC = 44^\circ$. Angle $CDE$ and Angle $ADC$ form a linear pair on line $AE$. So $\angle CDE = 180^\circ - 44^\circ = 136^\circ$. Let me re-evaluate standard properties. Actually, usually exterior angle equals interior opposite angle ONLY if defined that way. Let's stick to the basics. $\angle B = 44^\circ$. Opposite angle $\angle D$ (inside) $= 44^\circ$. $\angle CDE$ and $\angle D$ (inside) are supplementary. $180 - 44 = 136$. Let's check Angle $C$. $\angle C = 136^\circ$. $\angle C$ and $\angle CDE$ are alternate interior angles if we consider parallel lines $BC$ and $AD$ cut by transversal $CD$? No. $BC$ is parallel to $AD$. Transversal is $CD$. Angle $BCD$ and Angle $CDE$ are alternate interior angles? No. Let's use parallel lines $AB$ and $CD$. No. Let's use parallel lines $BC$ and $ADE$. Transversal $CD$. Angle $BCD$ and Angle $CDE$ are consecutive interior? No. Alternate Interior angles are formed by Z-shape. The Z is $B-C-D-E$? No. The Z is $B-C-D-A$. Angle $BCD$ and Angle $CDA$ are consecutive interior angles summing to 180. $136 + 44 = 180$. Correct. Now, line $ADE$ is a straight line. Angle $CDA + Angle CDE = 180$. $44 + Angle CDE = 180$. Angle $CDE = 136$.)*

*Correction*: Let's look at the position of $E$. It is on the extension of $AD$. So $A-D-E$ is a line.
$\angle B = 44^\circ$.
Opposite angle $\angle ADC = 44^\circ$.
$\angle ADC$ and $\angle CDE$ form a linear pair (straight line).
$\angle CDE = 180^\circ - 44^\circ = 136^\circ$.

*Wait, let me look at the diagram again.* Usually, these problems have a "trick". If $BC$ is parallel to $AD$, then $\angle B$ and $\angle A$ are supplementary. $\angle A = 136^\circ$. $\angle C = 136^\circ$. $\angle D_{internal} = 44^\circ$. $\angle CDE$ is adjacent to $\angle D_{internal}$. So $\angle CDE = 136^\circ$.

*Let's check another property:* Corresponding angles? If we extend $AB$ and $DC$, no. If we look at parallel lines $AB$ and $DC$? No. Parallel lines $BC$ and $AE$. Transversal $CD$. Angle $BCD$ and Angle $CDE$ are alternate interior angles? No. They are on the same side of the transversal? No.
Let's visualize the "Z". Go from $B$ to $C$, down to $D$, then to $E$. The angle inside the elbow at $C$ is $136^\circ$. The angle inside the elbow at $D$ is $\angle CDE$. Since $BC \parallel AD$ (and thus $BC \parallel DE$), the alternate interior angles are $\angle BCD$ and $\angle CDE$? No, that would require the transversal to connect them differently.
Actually, yes! If you extend line $BC$ to the right and line $AD$ to the right ($E$), and draw transversal $CD$, then $\angle BCD$ and $\angle CDE$ are alternate interior angles. Therefore, they are equal.
$\angle BCD = 136^\circ$. So $\angle CDE = 136^\circ$.

So, $m\angle CDE = 136^\circ$.

Problem 3

Given: Parallelogram $ABCD$ with diagonal $AC$. Angle $DAC$ is not given directly, but Angle $D = 83^\circ$ and Angle $ACD = 56^\circ$. Note: The label $83^\circ$ is at vertex $D$. The label $56^\circ$ is angle $ACD$.

1. Find angle $DCA$ ($m\angle DCA$): The diagram labels the angle between diagonal $AC$ and side $CD$ as $56^\circ$. So, $m\angle DCA = 56^\circ$. (This was given directly in the diagram).
2. Find angle $CAD$ ($m\angle CAD$): Look at triangle $\triangle ADC$. The sum of angles in a triangle is $180^\circ$.
We know $\angle D = 83^\circ$ and $\angle DCA = 56^\circ$.
$$\angle CAD = 180^\circ - 83^\circ - 56^\circ$$
$$180^\circ - 139^\circ = 41^\circ$$
So, $m\angle CAD = 41^\circ$.
3. Find angle $CBA$ ($m\angle CBA$): In a parallelogram, opposite angles are equal. Angle $CBA$ (angle $B$) is opposite to angle $D$ (angle $ADC$).
Since $\angle D = 83^\circ$, then $m\angle CBA = 83^\circ$.

Problem 4

Given: Parallelogram $ABCD$ with diagonals intersecting at $E$. Angle $AEB$ is not labeled, but Angle $CED$ is vertically opposite. Wait, the diagram shows Angle $CEB$? No, it shows Angle $DEC$? Let's look closely.
The angle labeled $110^\circ$ is $\angle CED$ (or $\angle AEB$ since they are vertical). Let's assume the angle inside the triangle $\triangle CDE$ at vertex $E$ is $110^\circ$? Or is it $\angle CEB$?
Looking at the arc, it is obtuse. It looks like $\angle CED = 110^\circ$ or $\angle AEB = 110^\circ$. Let's assume $\angle CED = 110^\circ$ based on typical positioning, or perhaps $\angle BEC$?
Actually, looking at the number 110, it is placed in the angle $\angle DEC$ (bottom triangle) or $\angle AEB$ (top)? It's in the left/right? It's in the angle $\angle DEC$? No, it's in $\angle CEB$?
Let's look at the other given: Angle $CAD$? No, Angle $DAC$? The angle at $A$ is split. The top part $\angle BAC$? No, the angle labeled $70^\circ$ is $\angle CAD$? Or $\angle ADB$?
The label $70^\circ$ is at vertex $A$, between diagonal $AC$ and side $AD$? No, it's between diagonal $BD$ and side $AD$? That would be $\angle ADB$. But the vertex is $A$. So it is $\angle CAD$ or $\angle BAC$. The arc is between side $AD$ and diagonal $AC$. So $\angle CAD = 70^\circ$?
Wait, if $\angle CAD = 70^\circ$ and $\angle D = ?$
Let's re-read the diagram carefully.
Vertex $A$: Angle between diagonal $AC$ and side $AB$? Or $AD$? The arc is near the top side $AB$? No, it's near the bottom side $AD$?
Let's look at Vertex $D$. Side $CD$ and Diagonal $BD$?
Let's look at the intersection $E$. The angle $110^\circ$ is $\angle CED$? Or $\angle AED$?
Usually, these diagrams label specific triangles.
Let's assume:
- $\angle CED = 110^\circ$ (Angle at intersection).
- $\angle CAD = 70^\circ$? No, that would make triangle $ADE$ sum $> 180$ if $\angle AED = 70$ (vert opp to 110? No, supp to 110 is 70).
If $\angle CED = 110^\circ$, then $\angle AED = 180 - 110 = 70^\circ$.
If the angle labeled $70^\circ$ at $A$ is $\angle CAD$, then in $\triangle ADE$: $\angle AED = 70^\circ$, $\angle DAE = 70^\circ$. Then $\angle ADE = 180 - 70 - 70 = 40^\circ$.
Then $\angle ABD = \angle BDC$ (alt interior).
Let's try another interpretation. Maybe the $110^\circ$ is $\angle AEB$? Then $\angle AED = 70^\circ$.
Maybe the $70^\circ$ is $\angle ADB$? No, vertex is $A$.
Let's assume the standard labeling:
- Angle at intersection $E$: $\angle DEC = 110^\circ$? Or $\angle AEB$? Vertical angles are equal. Let's assume the angle facing side $CD$ is $110^\circ$. So $\angle DEC = 110^\circ$.
- Angle at $A$: The arc is between diagonal $AC$ and side $AD$. So $\angle CAD = 70^\circ$.
- Length $ED = 9$.

Let's solve based on this:
1. Find angle $ECD$ ($m\angle ECD$):
First, find angles in $\triangle ADE$.
$\angle AED$ and $\angle DEC$ are supplementary. $\angle AED = 180^\circ - 110^\circ = 70^\circ$.
In $\triangle ADE$, we have $\angle DAE = 70^\circ$ (given as $\angle CAD$) and $\angle AED = 70^\circ$.
So $\triangle ADE$ is isosceles with $AD = ED$? No, base angles are at $A$ and $E$. So sides opposite are equal. Side opposite $\angle AED$ is $AD$. Side opposite $\angle DAE$ is $DE$. So $AD = DE = 9$.
Also, $\angle ADE = 180^\circ - 70^\circ - 70^\circ = 40^\circ$.

Now, we need $m\angle ECD$.
Since $AB \parallel CD$, alternate interior angles are equal.
$\angle BAC = \angle ACD$ (which is $\angle ECD$).
Do we know $\angle BAC$? Not yet.
Let's look at $\triangle CDE$.
We know $\angle DEC = 110^\circ$.
We need one more angle.
We know $\angle ADC = \angle ADE + \angle EDC$.
We found $\angle ADE = 40^\circ$.
We don't know $\angle EDC$ yet.
However, we know $AD \parallel BC$ and $AB \parallel CD$.
In parallelogram, opposite sides are equal. $AB = CD$, $AD = BC$.
We found $AD = 9$. So $BC = 9$.
Diagonals bisect each other. So $BE = ED = 9$. Thus $BD = 18$.
And $AE = EC$.

Let's go back to angles.
In $\triangle ADE$, angles are $70, 70, 40$.
So $\angle ADB = 40^\circ$.
Since $AD \parallel BC$, alternate interior angle $\angle DBC = \angle ADB = 40^\circ$.

What about $\angle ECD$?
Consider $\triangle CDE$.
We know $\angle DEC = 110^\circ$.
We need $\angle CDE$ or $\angle DCE$.
Note that $\angle CDE$ is part of angle $D$.
Angle $D$ (whole) = $\angle ADC$.
Angle $B$ (whole) = $\angle ABC$.
Opposite angles are equal.

Let's use the property that diagonals bisect each other.
$AE = EC$.
In $\triangle ADE$, we established it is isosceles with $AD = DE = 9$.
Since $DE = 9$, and diagonals bisect, $BE = 9$. So $BD = 18$.
Also $AD = 9$. Since opposite sides are equal, $BC = 9$.

Now, look at $\triangle BCE$.
$BC = 9$, $BE = 9$. So $\triangle BCE$ is isosceles.
Angle $\angle CBE = \angle DBC = 40^\circ$ (found above via alt interior with $\angle ADB$).
Since $\triangle BCE$ is isosceles with $BC=BE$, the base angles are $\angle BCE$ and $\angle BEC$? No, the vertex is $B$? No, $BC=BE$, so the angles opposite these sides are equal?
Sides $BC$ and $BE$ are equal. The angles opposite them are $\angle BEC$ and $\angle BCE$.
Wait, $BC=9$ and $BE=9$. So $\triangle BCE$ is isosceles.
The angle between the equal sides is $\angle CBE = 40^\circ$.
So the base angles are $(180 - 40) / 2 = 70^\circ$.
So $\angle BCE = 70^\circ$ and $\angle BEC = 70^\circ$.

Check consistency:
$\angle BEC$ and $\angle AED$ are vertical angles. We calculated $\angle AED = 70^\circ$. This matches $\angle BEC = 70^\circ$. Perfect.

Now, we need $m\angle ECD$.
$\angle ECD$ is the same as $\angle ACD$.
We know $\angle BCA = \angle DAC$ (alternate interior angles).
We were given $\angle DAC = 70^\circ$. So $\angle BCA = 70^\circ$.
From our calculation above, $\angle BCE$ (which is $\angle BCA$) is $70^\circ$. This is consistent.

We need $\angle ECD$.
We know consecutive angles sum to 180.
$\angle BCD + \angle ADC = 180^\circ$.
$\angle BCD = \angle BCE + \angle ECD = 70^\circ + \angle ECD$.
$\angle ADC = \angle ADE + \angle EDC = 40^\circ + \angle EDC$.
Also, alternate interior angles: $\angle ABD = \angle BDC$ (i.e., $\angle EDC$).
And $\angle BAC = \angle ACD$ (i.e., $\angle ECD$).

Let's find $\angle EDC$.
In $\triangle CDE$:
Sum of angles = 180.
$\angle DEC = 110^\circ$.
$\angle ECD + \angle EDC = 70^\circ$.

We also know that in parallelogram, opposite angles are equal.
$\angle DAB = \angle BCD$.
$\angle DAB = \angle DAC + \angle CAB = 70^\circ + \angle CAB$.
$\angle BCD = \angle BCA + \angle ACD = 70^\circ + \angle ACD$.
Since $\angle CAB = \angle ACD$ (alt interior), this is always true.

We need another relationship.
Look at $\triangle CDE$ and $\triangle ABE$. They are congruent.
Look at $\triangle ADE$ and $\triangle CBE$. They are congruent.

Is there information I missed?
Ah, look at side lengths.
$AD = 9$. $BC = 9$.
$DE = 9$. $BE = 9$.
So $BD = 18$.

Is the parallelogram a rhombus?
If $AD = DE = 9$, and $DE$ is half diagonal...
If diagonals are perpendicular, it's a rhombus. Here angles are 70/110, not 90. So not a rhombus.

Let's look at $\triangle CDE$ again.
We know $\angle DEC = 110^\circ$.
We need $\angle ECD$.
Do we know side $CD$?
In $\triangle ADE$, sides are $AD=9, DE=9$. Angle $D=40$.
Using Law of Cosines on $\triangle ADE$:
$AE^2 = 9^2 + 9^2 - 2(9)(9)\cos(40^\circ)$.
$AE^2 = 162 - 162(0.766) = 162(0.234) \approx 37.9$.
$AE \approx 6.16$.
So $EC = 6.16$.

In $\triangle CDE$:
Sides $DE=9, EC=6.16$. Angle $E=110$.
Use Law of Sines?
$\frac{CD}{\sin(110)} = \frac{EC}{\sin(\angle EDC)} = \frac{DE}{\sin(\angle ECD)}$.
We need $\angle ECD$. Let it be $x$.
Then $\angle EDC = 70 - x$.
$\frac{9}{\sin(x)} = \frac{6.16}{\sin(70-x)}$.
This seems too complex for this worksheet level. There must be a simpler geometric property.

Let's re-read the diagram numbers.
Maybe the $70^\circ$ is NOT $\angle CAD$.
Maybe the $70^\circ$ is $\angle ADB$? No, vertex A.
Maybe the $70^\circ$ is $\angle BAC$?
If $\angle BAC = 70^\circ$, then $\angle ACD = 70^\circ$ (alt interior).
Then in $\triangle CDE$:
$\angle ECD = 70^\circ$.
$\angle DEC = 110^\circ$.
Sum = $180^\circ$. Then $\angle EDC = 0^\circ$. Impossible.

So $\angle BAC$ cannot be 70.

What if the $110^\circ$ is $\angle AEB$?
Then $\angle AED = 70^\circ$.
If $\angle CAD = 70^\circ$, then $\triangle ADE$ has angles $70, 70, 40$. Same as before.

What if the $70^\circ$ is $\angle ADB$? (Label is near A, but maybe refers to the angle subtended by AB? No).

Let's look at the label "70" again. It is clearly at vertex A. Inside the triangle formed by diagonal AC and side AD? Or diagonal BD and side AD?
The line going from A is diagonal AC? Or BD?
In diagram 4, the diagonal from A goes to C. The diagonal from B goes to D.
The angle arc is between the top side (AB? No, A is top right? No, A is top right, B top left, C bottom left, D bottom right? Standard counter-clockwise?
Let's check labels.
Top Right: A. Top Left: B. Bottom Left: C. Bottom Right: D.
Diagonal AC connects Top Right to Bottom Left.
Diagonal BD connects Top Left to Bottom Right.
Intersection E.

Angle at A: The arc is between Side AB and Diagonal AC? Or Side AD and Diagonal AC?
Visually, the arc is acute. The angle $\angle DAB$ is likely obtuse or large acute.
The line segment involved is the diagonal going to C.
The other side of the angle is the side AD (bottom) or AB (top).
Given the orientation, the angle labeled 70 is likely $\angle CAD$ (between diagonal AC and side AD).

Angle at E: The label 110 is in the angle $\angle CED$? Or $\angle AEB$?
It is in the bottom triangle $\triangle CDE$? Or left $\triangle BCE$?
The number 110 is written in the space for $\angle CED$ (vertical to $\angle AEB$) or $\angle AED$?
It looks like it's in $\angle CED$ or $\angle AEB$. Let's assume $\angle CED = 110^\circ$ (obtuse).

If my previous calculation led to a complex sine rule, did I miss a simple isosceles triangle?
We found $\triangle ADE$ is isosceles ($AD=DE=9$).
We found $\triangle BCE$ is isosceles ($BC=BE=9$).

What about $\triangle CDE$?
Sides $DE=9$. $EC = AE$.
In $\triangle ADE$, $AE$ is the base.
Is it possible that $\triangle CDE$ is also isosceles?
If $EC = CD$? Or $DE = CD$?

Let's check the angles again.
$\angle CAD = 70$. $\angle ADE = 40$. $\angle AED = 70$.
$\angle ACB = 70$ (alt interior to CAD).
$\angle DBC = 40$ (alt interior to ADB).
$\angle BEC = 70$ (vertical to AED).
$\angle DEC = 110$ (supp to AED).

In $\triangle CDE$:
Angle $E = 110$.
Angle $C$ ($\angle ECD$) + Angle $D$ ($\angle EDC$) = 70.

We know $\angle BCD = \angle BCA + \angle ACD = 70 + \angle ECD$.
We know $\angle ADC = \angle ADB + \angle BDC = 40 + \angle EDC$.
Opposite angles equal: $\angle BCD = \angle DAB$.
$\angle DAB = \angle DAC + \angle CAB = 70 + \angle CAB$.
Since $\angle CAB = \angle ACD = \angle ECD$, this is consistent.

Consecutive angles supplementary:
$\angle DAB + \angle ADC = 180$.
$(70 + \angle ECD) + (40 + \angle EDC) = 180$.
$110 + \angle ECD + \angle EDC = 180$.
$\angle ECD + \angle EDC = 70$.
This just confirms the triangle sum. It doesn't give individual values.

Is there a constraint I missed?
"Find the values..." implies a unique solution.
Did I interpret the side length 9 correctly?
The 9 is on segment $ED$? Or $CD$?
The number 9 is written along the segment $ED$ (part of diagonal BD).

Is it possible the parallelogram is a Rhombus?
If it were a rhombus, diagonals would be perpendicular ($90^\circ$). Here we have 110/70. So no.
If it were a rectangle, diagonals equal.

Let's look at the label "70" again.
Could it be $\angle ADB$?
If $\angle ADB = 70^\circ$:
Then in $\triangle ADE$:
$\angle ADE = 70^\circ$.
$\angle AED = 70^\circ$ (if $\angle DEC=110$).
Then $\angle DAE = 180 - 70 - 70 = 40^\circ$.
So $\angle CAD = 40^\circ$.
Then $\triangle ADE$ is isosceles with $AD = AE$? No, angles at D and E are 70. So sides opposite are $AE$ and $AD$. So $AE = AD$.

Now look at $\triangle CDE$.
$\angle DEC = 110^\circ$.
$\angle CDE = \angle ABD$ (alt interior).
$\angle DCE = \angle BAC$ (alt interior).

If $\angle ADB = 70$, then $\angle DBC = 70$ (alt interior).
In $\triangle BCE$:
$\angle BEC = 70$ (vertical to AED).
$\angle CBE = 70$.
So $\triangle BCE$ has two 70s. So $\angle BCE = 180 - 140 = 40^\circ$.
So $\angle BCA = 40^\circ$.
Since $\angle BCA = \angle CAD$ (alt interior), then $\angle CAD = 40^\circ$.
This matches our deduction from $\triangle ADE$.

So this scenario is self-consistent.
Scenario A: Label 70 is $\angle CAD$. Result: $\triangle ADE$ isosceles ($AD=DE$).
Scenario B: Label 70 is $\angle ADB$. Result: $\triangle ADE$ isosceles ($AD=AE$).

Which one is it?
Visually, the arc for 70 is at vertex A. It is between the diagonal AC and the side AD.
Usually, an angle labeled at a vertex with an arc inside the corner refers to the angle formed by the two segments bounding the arc.
The segments are Side AD and Diagonal AC.
So $\angle CAD = 70^\circ$ is the most direct reading.

If $\angle CAD = 70^\circ$, we got stuck needing more info.
UNLESS... Look at the side labeled 9.
Is 9 the length of $CD$?
The number 9 is next to the segment $ED$? Or $CD$?
It is parallel to the diagonal $BD$. It is between E and D. So $ED = 9$.

Is it possible that $ABCD$ is a Rhombus?
If it's a rhombus, $AD = CD$.
If $AD = CD$, and we know $AD = 9$ (from Scenario A where $\triangle ADE$ isosceles $AD=DE=9$?? No, in Scenario A, $AD=DE=9$).
If $AD=9$ and $DE=9$, and it's a rhombus, then $CD=9$.
Then $\triangle CDE$ has sides $DE=9, CD=9$.
So $\triangle CDE$ is isosceles with $DE=CD$.
Then base angles $\angle DEC$? No, vertex is D? No, sides $DE$ and $CD$ meet at D.
So $\angle DEC = \angle DCE$? No.
Sides $DE=CD \implies \angle DEC = \angle DCE$? No.
Angles opposite equal sides are equal.
Angle opposite $CD$ is $\angle DEC = 110^\circ$.
Angle opposite $DE$ is $\angle DCE$.
So $\angle DCE = 110^\circ$.
Sum $110+110 > 180$. Impossible.

So it's not a rhombus with $DE=CD$.

What if $CE=CD$?

Let's reconsider the standard problems of this type.
Often, the "missing" piece is that the triangle is isosceles.
In Scenario A ($\angle CAD=70$), we found $\triangle ADE$ has angles $70-70-40$. So $AD=DE=9$.
We found $\triangle BCE$ has angles $70-70-40$. So $BC=BE=9$.

We need $\angle ECD$.
Is there any reason for $\triangle CDE$ to be isosceles?
If $ABCD$ is a parallelogram, $AB=CD$.
We don't know $AB$.

Wait! Look at Question 4 again.
$m\angle ECD = \_\_\_$
$m\angle AEB = \_\_\_$
$m\angle ABD = \_\_\_$
$BD = \_\_\_$

We can find $BD$ easily.
$BD = BE + ED$.
We know $ED = 9$.
In Scenario A, $\triangle BCE$ is isosceles with $BC=BE$.
We know $BC = AD$ (opp sides).
In $\triangle ADE$, $AD = DE = 9$.
So $BC = 9$.
So $BE = 9$.
So $BD = 9 + 9 = 18$.
This works perfectly.

Now, what about the angles?
$m\angle AEB$: Vertically opposite to $\angle CED$.
We assumed $\angle CED = 110^\circ$.
So $m\angle AEB = 110^\circ$.
(Or if the 110 was $\angle AEB$, then it's 110. If 110 was $\angle AED$, then $\angle AEB=70$. Given 110 is obtuse, and AEB looks obtuse in some drawings, but here AEB is top. CED is bottom. They are vertical. So they are equal. Let's assume the label 110 applies to the vertical pair AEB/CED or AED/BEC.
Looking at the arc for 110: It is in the angle $\angle CED$? Or $\angle AEB$?
Actually, usually the label is inside the angle. The number 110 is in the left angle $\angle AEB$? Or bottom $\angle CED$?
Let's assume $m\angle AEB = 110^\circ$ because it's explicitly asked for. If it was given, why ask?
Ah! If 110 is given for $\angle CED$, then asking for $\angle AEB$ is trivial (vertical angles).
If 110 is given for $\angle AED$, then $\angle AEB = 70$.
Let's look at the shape. Angle AEB is obtuse? Angle AED is acute?
In my Scenario A, $\angle AED = 70$ (acute) and $\angle AEB = 110$ (obtuse).
The diagram shows the angle at E labeled 110. The arc spans the obtuse angle.
So the obtuse angle is 110.
Therefore, $m\angle AEB = 110^\circ$ (and $\angle CED = 110^\circ$).
And $m\angle AED = 70^\circ$.

Now, $m\angle ECD$.
We established $\angle ECD + \angle EDC = 70^\circ$.
We need one of them.
Is it possible that $AC \perp BD$? No.
Is it possible that $AB = AD$? (Rhombus).
If Rhombus, diagonals bisect angles.
$\angle CAD = \angle BAC$.
If $\angle CAD = 70$, then $\angle BAC = 70$.
Then $\angle DAB = 140$.
Then $\angle ADC = 40$.
Then $\angle ADB = 20$.
But we calculated $\angle ADB = 40$ in Scenario A. Contradiction.
So not a rhombus.

Let's look at the label "70" again.
What if the 70 is $\angle ABD$?
No, it's at A.

What if the 70 is $\angle BAC$?
If $\angle BAC = 70$, then $\angle ACD = 70$.
Then $\angle ECD = 70$.
Then in $\triangle CDE$: $\angle E = 110, \angle C = 70$. Sum = 180. $\angle D = 0$. Impossible.

What if the 70 is $\angle ADB$? (Misplaced label?)
If $\angle ADB = 70$:
Then $\angle DBC = 70$.
$\angle AED = \angle BEC$.
If $\angle AEB = 110$, then $\angle AED = 70$.
In $\triangle ADE$: $\angle ADE = 70, \angle AED = 70 \implies \angle DAE = 40$.
So $\angle CAD = 40$.
Then $\angle ACB = 40$.
In $\triangle BCE$: $\angle CBE = 70, \angle BCE = 40 \implies \angle BEC = 70$.
Matches $\angle AED = 70$.
So this is a consistent geometry.
In this case:
$m\angle ECD$?
$\angle BCD = \angle BCA + \angle ACD = 40 + \angle ACD$.
$\angle ADC = \angle ADB + \angle BDC = 70 + \angle BDC$.
Opposite angles: $\angle BCD = \angle DAB = \angle DAC + \angle CAB = 40 + \angle CAB$.
Alt interior: $\angle CAB = \angle ACD$.
So $\angle BCD = 40 + \angle ACD$. Consistent.
Consecutive: $\angle DAB + \angle ADC = 180$.
$(40 + \angle ACD) + (70 + \angle BDC) = 180$.
$\angle ACD + \angle BDC = 70$.
In $\triangle CDE$: $\angle E = 110$. Sum other two = 70. Consistent.

Does this give a unique value for $\angle ECD$?
No. $\angle ECD$ could be anything as long as $\angle EDC = 70 - \angle ECD$.

THEREFORE: There must be a visual cue I am missing or a standard assumption.
Look at side $CD$ and $DE$.
In the "70 is $\angle ADB$" case:
$\triangle ADE$ is isosceles ($AE=AD$).
$\triangle BCE$ is isosceles ($BE=BC$? No. $\angle BEC=70, \angle CBE=70 \implies CE=CB$).
So $CE = CB$.
We know $CB = AD$.
So $CE = AD$.
And $AD = AE$.
So $AE = CE$. (Diagonals bisect, so this is always true).

Is there a side length given for CD? No.

Let's go back to Scenario A ($\angle CAD = 70$).
Why would $\angle ECD$ be solvable?
Maybe $\triangle CDE$ is isosceles?
If $CD = DE = 9$?
If $CD = 9$, then $\triangle CDE$ has sides $9, 9, EC$.
Then $\angle DCE = \angle DEC$? No, $\angle DCE = \angle CDE$?
If $CD=DE$, angles opposite are $\angle DEC$ (110) and $\angle DCE$.
So $\angle DCE = 110$. Impossible.

If $CE = DE = 9$?
Then $\angle ECD = \angle EDC$.
Sum = 70. So each is 35.
$m\angle ECD = 35^\circ$.

If $CE = CD$?
Then $\angle CDE = \angle CED = 110$. Impossible.

So, is $CE = DE$?
$DE = 9$.
$CE = AE$.
In Scenario A, $\triangle ADE$ has sides $9, 9, AE$.
$AE = 2 \cdot 9 \cdot \sin(20^\circ)$? No.
Base $AE$. Angles 70-70-40.
$AE / \sin(40) = 9 / \sin(70)$.
$AE = 9 \sin(40) / \sin(70) \approx 9(0.64)/0.94 \approx 6.1$.
So $CE \approx 6.1$.
$DE = 9$.
So $CE \neq DE$.

Okay, look at the diagram 4 again.
Is the "70" actually $\angle ACD$?
If $\angle ACD = 70$, then $\angle ECD = 70$.
Then $\angle EDC = 0$. No.

Is the "70" actually $\angle BDC$?

Let's guess the intended answer based on common integer angles.
If $\angle ECD = 35^\circ$, it implies isosceles $\triangle CDE$ with $CE=DE$.
This happens if $AE=DE$.
In $\triangle ADE$, if $AE=DE$, then $\angle DAE = \angle ADE$.
$\angle AED = 70$ (if 110 is obtuse vert).
Then $2x + 70 = 180 \implies 2x=110 \implies x=55$.
So $\angle CAD = 55^\circ$.
But label says 70.

What if the 110 is $\angle AED$?
Then $\angle AEB = 70$.
Then $\angle DEC = 70$.
Then $\angle AEB = 70$.
If $\angle CAD = 70$:
$\triangle ADE$: $\angle AED = 110$. $\angle DAE = 70$. Sum > 180. Impossible.

So 110 MUST be the obtuse angle $\angle AEB / \angle CED$.

Final check on Problem 4.
If the question asks for $BD$, and $BD=18$ is derived from $AD=DE=9$, this relies on $\triangle ADE$ being isosceles with $AD=DE$.
This requires $\angle DAE = \angle AED$.
$\angle AED = 70$.
So $\angle DAE$ must be 70.
The label 70 is at A.
So $\angle CAD = 70$.
This confirms Scenario A is the intended path.

So, we have:
$BD = 18$.
$m\angle AEB = 110^\circ$.

Remaining: $m\angle ECD$ and $m\angle ABD$.