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

$$\begin{gathered} \mathrm{In}\mathrm{\Delta AOL}\mathrm{and}\mathrm{\Delta CMO}: \\ \angle \mathrm{AOL}=\angle \mathrm{COM}(v\mathrm{ertically}\mathrm{opposite}\mathrm{angle})....\left(i\right) \\ \angle \mathrm{ALO}=\angle \mathrm{CMO}=90°(e\mathrm{ach}\mathrm{right}\mathrm{angle}).....\left(ii\right) \\ \mathrm{Usin}ganglesumproperty: \\ \angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle LAO=180°..........\left(iii\right) \\ \angle COM+\angle CMO+\angle OCM=180°......\left(iv\right) \\ Fromequations\left(iii\right)and\left(iv\right): \\ \angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle LAO=\angle COM+\angle CMO+\angle OCM \\ \angle LAO=\angle OCM(fromequations(i)and(ii)) \\ \mathrm{In}\mathrm{\Delta AOL}\mathrm{and}\mathrm{\Delta CMO}: \\ \angle \mathrm{ALO}=\angle \mathrm{CMO}(e\mathrm{ach}\mathrm{right}\mathrm{angle}) \\ \mathrm{AO}=\mathrm{OC}(\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{parallelogram}\mathrm{bisect}\mathrm{each}\mathrm{other}) \\ \angle LAO=\angle OCM(provedabove) \\ \mathrm{So},\mathrm{\Delta AOL}\mathrm{is}\mathrm{congruent}\mathrm{to}\mathrm{\Delta CMO}\left(S\mathrm{AS}\right). \\ \Rightarrow \mathrm{AL}=\mathrm{CM}\left[\mathrm{cpct}\right] \end{gathered}$$