Question

In the figure , seg BD perpendicular side AC , seg DE perpendicular side BC , then show that DE× BE = 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△BED and △BDC

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

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

So, \triangle BED \sim \triangle BDC△BED∼△BDC    by AA similarity

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

\dfrac{DE}{DC}=\dfrac{BE}{BD}DCDE​=BDBE​

Using cross multiply

DExBD=DCxBE

Hence Proved

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