Question

in the fig seg BD perpendicular to side AC seg DE perpendicular to side BC then show that DE×BD=DC×BE

Answer 1

Given: BD perpendicular to side AC ( BD⊥AC) and DE perpendicular to side BC ( DE⊥BC)

To prove: DExBD=DCxBE

Proof: In $\triangle BED\text{ and }\triangle BDC$

$\angle BED=\angle BDC$           (Each 90° )

$\angle DBE=\angle DBC$           (Common in both triangle)

So, $\triangle BED \sim \triangle BDC$    by AA similarity

If two triangles are similar then their corresponding sides are in proportional.

$\dfrac{DE}{DC}=\dfrac{BE}{BD}$

Using cross multiply

DExBD=DCxBE

Hence Proved

Answer 2

Hence proved. Hope it help you