ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig.) . Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ

Answer:

Given, ABCD is a parallelogram.

     BD is one of its diagonal and AP & CQ are perpendiculars on BD.

  Now, we have ΔAPB and ΔCQD.

  We have, AB = CD [Opposite sides of a parallelogram]

  ∠ABP = ∠CDQ [Since, AB ∥ CD and BD is the bisector, hence ∠ABP & ∠CDQ are alternate angles]

     ∠APB = ∠CQD [Since, both are right angles]

     Therefore, by Angle-Side-Angle theorem, ΔAPB ≅ ΔCQD …………. [Proved (i)]

     With similar argument, since ΔAPB ≅ ΔCQD, we can easily conclude that AP = CQ [Similar sides of two similar triangles are of same length] …………… [Proved (ii)]