Q) Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. 

Prove that:
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

(Q 34 B – 30/5/2 – CBSE 2026 Question Paper)

Ans:

(i) Prove that AQ = QR

Step 1: Given that ABCD is a parallelogram

∴ AB = CD

∵ Q is mid point of side CD 

∴ DQ = QC

∵ CD = DQ + QC

∴ CD = QC + QC              (∵ DQ = QC)

∴ CD = 2 QC

∴ AB = 2 QC ….. (i)            (∵ AB = CD)

Step 2: Now, in Δ ARB, QC ǁ AB (∵ CD ǁ AB)

∴ By BPT or Basic Proportionality Theorem (Thales’s Theorem),

∴          

(∵ AB = 2 QC proved above)

∴

∴ AR = 2 QR

Step 3: ∵ AR = AQ + QR

∴ AQ + QR = 2 QR

∴ AQ = QR……. Hence Proved!

(ii) Prove that AP = 2PQ

Step 4: Let’s compare Δ APB and Δ QPD

∠ DQP = ∠ PAB (CD ǁ AB, and AQ cuts these lines)

∠ APB = ∠ QPD (Alternate angles)

∴ Δ APB ~ Δ QPD

Step 5: ∴ …….. (iii)

Since AB = CD and CD = CQ + DQ

∴ AB = CQ + DQ

∵ CQ = DQ     (Q is mid point of CD)

∴ AB = 2 DQ

Substituting this value in equation (iii), we get:

∴

∴ = 2

∴ AP = 2 PQ…….Hence Proved!

(iii) Prove that PR = 2AP

Step 6: From part (i), we proved that AQ = QR

let’s add PQ on both sides

∴ AQ + PQ = QR + PQ

∴ AQ + PQ = PR                   (∵ QR + PQ = PR)

∴ (AP + PQ) + PQ = PR        (∵ AP + PQ = AQ)

∴ AP + 2 PQ = PR

∵ AP = 2 PQ from part (ii)

∴ AP + AP = PR 

∴ 2 AP = PR

or PR = 2 AP ……..Hence Proved !

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