Question

p is the mid point of the side AD of a parallelogram ABCD. Straight line BP intersects the diagonal AC at R and the side CD (produced) at Q. Prove QR=2RB​

Answer 1

Given: P is the mid point of the side AD of a parallelogram ABCD. Straight line BP intersects the diagonal AC at R and the side CD (produced) at Q.

To find: Prove QR=2RB​.

Solution:

• Consider the triangle QPD and triangle PAB, we have:

                   ∠QDP = ∠PAB  (Pair of alternate angles)

                   DP = PA  (P is the midpoint of AD)

                   ∠DPQ = ∠APB  (Vertically opposite angles)

• So by ASA congruence criterion, we have:

                   triangle QPD ≅ triangle PAB

                   QD = AB    (C.P.C.T.)

• Now, consider the triangle QRC and triangle ARB, we have:

                   ∠QRC = ∠ARB  (vertically opposite angles)

                   ∠CQR = ∠RBA  (pair of alternate angles)

• So by AA similarity criterion, we have:

                   triangle  QRC ≅ triangle ARB

• Now according to corresponding sides of similar triangle, we get:

                   QC/AB = CR/RA = QR/RB

                   QR/RB = QC/AB

                   QR/RB = QD + DC / AB

                   QR/RB = CD + DC / AB   ..........(as QD = AB and AB = CD )

                   QR/RB = 2CD / CD

                   QR/RB = 2

                   QR = 2 RB

• Hence proved.

Answer:

                So by similarity criteria and congurency criteria we proved that QR = 2 RB