Diagonals of a parallelogram ABCD intersect at O.

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


In $\Delta \mathrm{AOL}$ and $\Delta \mathrm{CMO}:$

$\angle \mathrm{AOL}=\angle \mathrm{COM}($ vertically opposite angle $) \ldots(i)$

$\angle \mathrm{ALO}=\angle \mathrm{CMO}=90^{\circ}(e$ ach right angle $) \ldots($ ii $)$

Usin $g$ angle sum property $:$

$\angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle L A O=180^{\circ} \ldots \ldots \ldots($ iii $)$

$\angle C O M+\angle C M O+\angle O C M=180^{\circ} \ldots \ldots($ iv $)$

From equations $(i i i)$ and $(i v)$ :

$\angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle L A O=\angle C O M+\angle C M O+\angle O C M$

$\angle L A O=\angle O C M$ (from equations $(i)$ and $(i i))$

In $\Delta \mathrm{AOL}$ and $\Delta \mathrm{CMO}:$

$\angle \mathrm{ALO}=\angle \mathrm{CMO}(e$ ach right angle $)$

$\mathrm{AO}=\mathrm{OC}$ (diagonals of a parallelogram bisect each other)

$\angle L A O=\angle O C M$ (proved above)

So, $\Delta \mathrm{AOL}$ is congruent to $\Delta \mathrm{CMO}(S \mathrm{AS})$

$\Rightarrow \mathrm{AL}=\mathrm{CM}[\mathrm{cpct}]$


Leave a comment

Please enter comment.
Please enter your name.