Quadrilaterals Worksheets - Math Monks - Free Printable

Let’s solve each problem step by step, using properties of parallelograms and rectangles. Since you uploaded a worksheet titled “Theorems Involving Quadrilaterals,” we’ll rely on standard geometric theorems.

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Problem 1: PQRS is a parallelogram. Prove that (a) PQ ≅ RS, (b) PS ≅ RQ

Given: PQRS is a parallelogram → By definition, opposite sides are parallel:
→ PQ ∥ RS and PS ∥ RQ

To prove: Opposite sides are congruent.

Proof:

Draw diagonal PR.

In ΔPQR and ΔRSP:
- PQ ∥ RS → ∠QPR ≅ ∠SRP (Alternate Interior Angles)
- PS ∥ RQ → ∠QRP ≅ ∠SPR (Alternate Interior Angles)
- PR ≅ PR (Reflexive Property)

∴ ΔPQR ≅ ΔRSP by ASA (Angle-Side-Angle)

Therefore, corresponding sides are congruent:
→ PQ ≅ RS (part a)
→ PS ≅ RQ (part b)

✔ Proved

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Problem 2: JKLM is a parallelogram. Prove that ∠J ≅ ∠L and ∠M ≅ ∠K

Given: JKLM is a parallelogram → JK ∥ LM, JM ∥ KL

To prove: Opposite angles are congruent.

Proof:

Draw diagonal JL.

In ΔJKL and ΔLMJ:
- JK ∥ LM → ∠KJL ≅ ∠MLJ (Alternate Interior Angles)
- JM ∥ KL → ∠K LJ ≅ ∠MJL (Alternate Interior Angles)
- JL ≅ JL (Reflexive)

∴ ΔJKL ≅ ΔLMJ by ASA

Therefore, ∠J ≅ ∠L (corresponding angles in congruent triangles)

Similarly, draw diagonal KM to show ∠M ≅ ∠K.

Alternatively, use the fact that consecutive angles are supplementary:

∠J + ∠K = 180° (consecutive angles in parallelogram)
∠K + ∠L = 180°
⇒ ∠J = ∠L (by substitution)

Same logic for ∠M ≅ ∠K.

✔ Proved

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Problem 3: EFGH is a parallelogram. Prove that ∠E and ∠F are supplementary

Given: EFGH is a parallelogram → EF ∥ HG, EH ∥ FG

To prove: ∠E + ∠F = 180°

Proof:

Since EF ∥ HG and EH is a transversal,
→ ∠E and ∠F are consecutive interior angles (same-side interior angles).

By the Same-Side Interior Angles Theorem, if two parallel lines are cut by a transversal, then the same-side interior angles are supplementary.

Therefore, ∠E + ∠F = 180°

✔ Proved

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Problem 4: ABCD and CDRS are parallelograms. Prove: AB ≅ SR

Given: ABCD is a parallelogram → AB ∥ CD and AB ≅ CD
CDRS is a parallelogram → CD ∥ SR and CD ≅ SR

To prove: AB ≅ SR

Proof:

From ABCD: AB ≅ CD (opposite sides of parallelogram)

From CDRS: CD ≅ SR (opposite sides of parallelogram)

By Transitive Property of Congruence:
AB ≅ CD and CD ≅ SR ⇒ AB ≅ SR

✔ Proved

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Problem 5: UVWX is a rectangle. Prove: (a) UV ∥ WX, (b) UW ∥ VX

Given: UVWX is a rectangle → All angles are 90°, and it is a special type of parallelogram.

To prove: Opposite sides are parallel.

Proof:

(a) UV ∥ WX

Since UVWX is a rectangle, it is a parallelogram by definition (a rectangle is a parallelogram with four right angles). In any parallelogram, opposite sides are parallel.

Thus, UV ∥ WX.

(b) UW ∥ VX

Similarly, UW and VX are the other pair of opposite sides → also parallel by definition of parallelogram.

✔ Proved

*(Note: In a rectangle, adjacent sides are perpendicular, but opposite sides remain parallel — this is inherent in the definition.)*

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Problem 6: ABCD is a parallelogram. Prove: (a) AB ≅ DC, (b) AD ≅ BC

This is essentially the same as Problem 1 — proving opposite sides are congruent.

Given: ABCD is a parallelogram → AB ∥ DC, AD ∥ BC

Proof:

Draw diagonal AC.

In ΔABC and ΔCDA:
- AB ∥ DC → ∠BAC ≅ ∠DCA (Alternate Interior Angles)
- AD ∥ BC → ∠BCA ≅ ∠DAC (Alternate Interior Angles)
- AC ≅ AC (Reflexive)

∴ ΔABC ≅ ΔCDA by ASA

Therefore:
→ AB ≅ DC (part a)
→ AD ≅ BC (part b)

✔ Proved

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## ✔ Final Summary of All Proofs:

| Problem | What to Prove | Key Theorem Used |
|--------|----------------|------------------|
| 1 | Opposite sides congruent | Triangle congruence (ASA) via diagonal |
| 2 | Opposite angles congruent | Triangle congruence or angle supplement property |
| 3 | Consecutive angles supplementary | Same-side interior angles theorem |
| 4 | AB ≅ SR via shared side CD | Transitive property of congruence |
| 5 | Opposite sides parallel | Rectangle ⇒ parallelogram ⇒ opposite sides parallel |
| 6 | Opposite sides congruent | Triangle congruence (ASA) |

These proofs rely on foundational theorems about parallelograms and rectangles. Remember: All rectangles are parallelograms, so they inherit all parallelogram properties.

Let me know if you’d like diagrams or step-by-step constructions for any proof!