Question

Using the conditions from 32(b), prove that \(AP = 2PQ\).

Hint:

Express all segments in terms of one major segment (like \(AQ\)) to compare them easily.

Answer 1

Step 1: Using Similarity of Triangles:
Consider triangles APD and RPB.

Since AD ∥ BR,
∠PAD = ∠PRB (alternate interior angles)
∠PDA = ∠PBR (alternate interior angles)

Therefore,
ΔAPD ∼ ΔRPB (AA similarity)

Step 2: Writing Proportional Sides:
From similarity,
AP / PR = AD / BR

In parallelogram ABCD,
AD = BC

Also from given construction and part (i),
CR = AD

Hence,
BR = BC + CR
= AD + AD
= 2AD

Substitute in ratio:
AP / PR = AD / 2AD
AP / PR = 1/2

So,
PR = 2AP

Step 3: Express AR in Terms of AP:
AR = AP + PR
= AP + 2AP
= 3AP

Given Q is midpoint of AR,
AQ = AR/2
= (3AP)/2

So,
AP = (2/3)AQ

Step 4: Finding PQ:
PQ = AQ − AP
= AQ − (2/3)AQ
= (1/3)AQ

Step 5: Comparing AP and PQ:
AP = (2/3)AQ
PQ = (1/3)AQ

Therefore,
AP = 2PQ

Final Answer:
AP = 2PQ