Q6 of 49 Page 151

Let ABCD be a square and let points P on AB and Q on DC be such that DP = AQ. Prove that BP = CQ.


Consider ΔADQ and ΔDAP


DP = AQ … given


∠PAD = ∠QDA … angles of square both 90°


AD = AD … common side


Therefore, by SAS test for congruency


ΔADQ ≅ ΔDAP


⇒ AP = DQ … corresponding sides of congruent triangles (i)


Now AB and CD are sides of square therefore,


AB = CD


⇒ AP + PB = DQ + QC … from figure AB = AP + PB and CD = DQ + QC


But using (i) AP = DQ


⇒ DQ + PB = DQ + QC


⇒ PB = QC


⇒ BP = CQ


Hence proved


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