Question

DB perpendicular BC, DE perpendicular AB and AC perpendicular BC.
Prove that BE/DE= AC/BC
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Answer 1

Answer:

In triangle BED and triangle ACB

Angle BED=angle ACB=90°

since, angle B+angle C=180°

therefore, BD||AC

angle EBD=angle CAB(alternate angles)

therefore, by AA similarity

triangle BED~triangle ACB

BE/AC=DE/BC

therefore, BE/DE=AC/BC