Geometry worksheet featuring Regents Exam questions focused on proving properties of quadrilaterals, including parallelograms and congruent triangles.

1. 3) Corresponding parts of congruent triangles are congruent.
2. Proof:
- Given: Parallelogram ABCD with diagonal AC drawn.
- Since ABCD is a parallelogram, AB ≅ CD and AD ≅ BC (opposite sides of a parallelogram are congruent).
- AC ≅ AC (Reflexive Property of Congruence).
- Therefore, △ABC ≅ △CDA by SSS (Side-Side-Side) Congruence Postulate.
3. Proof:
- Given: Quadrilateral ABCD, diagonal AFEC, AE ≅ FC, BF ⊥ AC, DE ⊥ AC, ∠1 ≅ ∠2.
- Since BF ⊥ AC and DE ⊥ AC, ∠BFA ≅ ∠DEC ≅ 90° (definition of perpendicular lines).
- In △ABF and △CDE: ∠1 ≅ ∠2 (given), ∠BFA ≅ ∠DEC (both right angles), and AE ≅ FC implies AF ≅ CE (since AE + EF = AF and CF + FE = CE, and EF is common, so AF = CE if AE = FC).
- Actually, to correct the logic: We have AE ≅ FC, and since E and F are on AC, we can say AF ≅ CE only if we assume E and F are positioned such that the segments add up correctly. However, the given AE ≅ FC and the perpendiculars suggest we look at triangles AFB and CED.
- Let’s reframe: Consider triangles AFB and CED.
- ∠1 ≅ ∠2 (given).
- ∠AFB ≅ ∠CED (both 90°).
- We need another side. Note that AE ≅ FC, but that doesn’t directly give us AF ≅ CE unless we know more about positions. Alternatively, perhaps the intention is to use ASA or AAS.
- Actually, let's consider triangles BFC and DEA? Not directly.
- Better approach: Use the given to show AB || CD and AD || BC.
- Since BF ⊥ AC and DE ⊥ AC, then BF || DE (two lines perpendicular to the same line are parallel).
- Now, in triangles ABF and CDE:
- ∠BAF ≅ ∠DCE? Not necessarily.
- Alternative: Use the fact that AE ≅ FC and the right angles to show that triangles AED and CFB are congruent? Not quite.
- Let’s try this: Since BF ⊥ AC and DE ⊥ AC, then BF || DE.
- Also, AE ≅ FC.
- Consider triangles AFE and CEF? Not helpful.
- Perhaps the key is to show that AB ≅ CD and AD ≅ BC or that opposite sides are parallel.
- Given ∠1 ≅ ∠2, and since BF and DE are both perpendicular to AC, then in triangles ABF and CDE:
- ∠1 ≅ ∠2 (given).
- ∠AFB ≅ ∠CED (both 90°).
- If we can show AF ≅ CE, then by AAS, △ABF ≅ △CDE.
- But AF = AE + EF, and CE = CF + FE. Since AE ≅ FC, and EF is common, then AF ≅ CE.
- So, △ABF ≅ △CDE by AAS.
- Therefore, AB ≅ CD and ∠BAF ≅ ∠DCE.
- Since ∠BAF ≅ ∠DCE, and they are alternate interior angles for lines AB and CD with transversal AC, then AB || CD.
- Similarly, we can show AD || BC by considering other triangles or using the same logic.
- Since AB || CD and AB ≅ CD, then ABCD is a parallelogram (one pair of opposite sides both parallel and congruent).
4. Proof:
- Given: Quadrilateral ABCD, AD ≅ BC and ∠DAE ≅ ∠BCE. Line segments AC, DB, and FG intersect at E.
- We are to prove: △AEF ≅ △CEG.
- Note: The diagram shows points F and G on AB and CD respectively, and FG passes through E.
- From the given, AD ≅ BC and ∠DAE ≅ ∠BCE.
- Also, ∠AED ≅ ∠CEB (vertically opposite angles).
- Therefore, in triangles ADE and CBE:
- AD ≅ BC (given).
- ∠DAE ≅ ∠BCE (given).
- ∠AED ≅ ∠CEB (vertically opposite angles).
- So, △ADE ≅ △CBE by ASA (Angle-Side-Angle).
- Therefore, AE ≅ CE and DE ≅ BE.
- Now, consider triangles AEF and CEG.
- AE ≅ CE (from above).
- ∠AEF ≅ ∠CEG (vertically opposite angles).
- We need another pair. Note that ∠FAE ≅ ∠GCE? Not directly given.
- Alternatively, since we have AE ≅ CE and ∠AEF ≅ ∠CEG, if we can show ∠AFE ≅ ∠CGE, then by AAS.
- But we don't have that.
- Perhaps from the congruence of △ADE and △CBE, we have ∠ADE ≅ ∠CBE.
- And since AD ≅ BC and AB and CD are sides, perhaps we can show that AB || CD? Not necessarily.
- Another approach: Since AE ≅ CE and E is on AC, and FG passes through E, and if we assume that F and G are midpoints or something, but not given.
- Actually, the problem might intend for us to use the fact that ∠DAE ≅ ∠BCE and the vertical angles to get the triangle congruence for AEF and CEG.
- Let’s look at angles at E: ∠AEF and ∠CEG are vertically opposite, so congruent.
- Also, from △ADE ≅ △CBE, we have ∠DAE ≅ ∠BCE, which are the same as ∠FAE and ∠GCE if F is on AB and G on CD.
- So, in △AEF and △CEG:
- ∠FAE ≅ ∠GCE (given as ∠DAE ≅ ∠BCE, and assuming F and G are on AB and CD so that these angles are the same).
- AE ≅ CE (from △ADE ≅ △CBE).
- ∠AEF ≅ ∠CEG (vertically opposite angles).
- Therefore, △AEF ≅ △CEG by ASA.
5. Proof:
- Given: Parallelogram FLSH, diagonal FGAS, LG ⊥ FS, HA ⊥ FS.
- Prove: △LGS ≅ △HAF.
- Since FLSH is a parallelogram, FL ≅ SH and FH ≅ LS, and FL || SH, FH || LS.
- Diagonal FGAS means that F, G, A, S are colinear, with G and A on the diagonal.
- LG ⊥ FS and HA ⊥ FS, so LG || HA (both perpendicular to FS).
- Also, since FLSH is a parallelogram, FS is one side, and LH is the opposite side, so FS || LH.
- Now, consider triangles LGS and HAF.
- First, note that ∠LGS and ∠HAF are both right angles? Not necessarily, because LG ⊥ FS and HA ⊥ FS, but G and A are on FS, so in triangle LGS, ∠LGS is the angle at G, which is between LG and GS. Since LG ⊥ FS and GS is part of FS, then ∠LGS = 90°. Similarly, ∠HAF = 90° because HA ⊥ FS and AF is part of FS.
- So, ∠LGS ≅ ∠HAF (both 90°).
- Also, since FLSH is a parallelogram, LS ≅ FH (opposite sides).
- Now, we need another side or angle.
- Consider the diagonal: since FGAS is the diagonal, and G and A are points on it, but we don't know if they are the same point or not.
- Actually, the diagonal is probably FS, and G and A are feet of perpendiculars from L and H to FS.
- So, LG and HA are altitudes to FS.
- In parallelogram FLSH, the area can be expressed as base FS times height, and since LG and HA are both heights to FS, and the parallelogram has equal area, but also, because FL || SH, the distance between them is constant, so LG = HA (the height is the same).
- So, LG ≅ HA.
- Now, in triangles LGS and HAF:
- LG ≅ HA (as established, both are heights to the same base in a parallelogram, so equal).
- ∠LGS ≅ ∠HAF (both 90°).
- We need another side. Note that GS and AF are parts of FS.
- Since FLSH is a parallelogram, and LG and HA are perpendiculars to FS, then the segments FG and AS might be related.
- Actually, consider that triangles FLG and HSA might be congruent, but let's focus.
- We have LG ≅ HA, ∠LGS ≅ ∠HAF, and if we can show GS ≅ AF, then by SAS.
- How to show GS ≅ AF?
- Note that in parallelogram FLSH, if we drop perpendiculars from L and H to FS, since FL || SH, and FS is the base, then the feet G and A should be such that FG = AS, because the parallelogram is symmetric in that way.
- Specifically, the distance from F to G should equal the distance from S to A, because the projection of FL and SH onto FS should be equal.
- Since FL ≅ SH and they are parallel, the horizontal components (along FS) should be equal.
- Therefore, FG = AS.
- Then, since FS = FG + GA + AS, and if FG = AS, then GS = GA + AS = GA + FG, and AF = AG + GF = AG + FG, so GS = AF.
- Thus, GS ≅ AF.
- Therefore, in △LGS and △HAF:
- LG ≅ HA (heights).
- ∠LGS ≅ ∠HAF (right angles).
- GS ≅ AF (as shown).
- So, △LGS ≅ △HAF by SAS.