Question

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

Answer 1

In parallelogram ABM diagonals AC and BD intersect each other at O.

$\therefore$ Ois the mid-point of AC and BD.
al $\perp BD$ and $CM\perp BD$.
In $\Delta ALO$ and $\Delta CMO$
$\angle L=\angle M$ $\left(\text{Each}{90}^{\circ }\right)$
$\angle AOL=\angle COM$
(Vertically opposite angles)
AO = CO ($\because$ O is mid-point of AC)
$\therefore \Delta ALO\cong \Delta COM$ (AAS axion)
$\therefore$ AL = CM (c.p.c.t)
Hence proved