Diagonals of a parallelogram `A B C D` intersect at `OdotA L\ a n d\ C M` are perpendiculars to `B D` such that `L\ a n d\ M` lie on `B D` . IS `A L=C M ?` why or why not?

Explanation

Giventhat
` AL `and `CM `are perpendiculars on diagonal `BD`
In `triangleAOL` and `triangleCOM`
`angleAOL = anglrCOM` (vertically opposite angle) `….. (i) `
`angleALO = angleCMO = 90^0` (each right angle) `……. (ii) `
using angle sum property `angleAOL + angleALO + angleLAO = 180^0 ……… (iii)`
`∠COM + ∠CMO + ∠OCM = 180° ……. (iv)`
From (iii) and (iv)
`angleAOL + angleALO + angleLAO = angleCOM + angleCMO + angleOCM`
`angleLAO = angleOCM`
(from (i) and (ii)) In `triangleAOL` and `triangleCOM`
`angleALO = angleCMO` (each right angle)
`AO = OC` (diagonals of a parallelogram bisect each other)
`angleLAO = angleOCM `(proved)
So
`triangleAOL` is congruent to `triangleCOM `
`AL = CM`(Corresponding parts of congruent triangles)