Question

Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM? Why or why not?

Answer 1

Given, AL and CM are perpendiculars on diagonal BD.

In ΔAOL and ΔCOM

∠AOL = ∠COM (vertically opposite angle) ….. (i)

∠ALO = ∠CMO = 90° (each right angle) ……. (ii)

By using angle sum property

∠AOL + ∠ALO + ∠LAO = 180° ……… (iii)

∠COM + ∠CMO + ∠OCM = 180° ……. (iv)

From (iii) and (iv)

∠AOL + ∠ALO + ∠LAO = ∠COM + ∠CMO + ∠OCM

∠LAO = ∠OCM (from (i) and (ii))

In ΔAOL and ΔCOM

∠ALO = ∠CMO (each right angle)

AO = OC (diagonals of a parallelogram bisect each other)

∠LAO = ∠OCM (proved)

So, ΔAOL is congruent to ΔCOM

∴ AL = CM (Corresponding parts of congruent triangles)