Question

two triangles ABC and DBC are on the same base BC in which angle A=B=90°. if CA and BD meet each other at E ,show that AE×CE=BE×DE.

Answer 1

proof: in ∆ ABC and DBC,
angle1=a4 ......each 90°
angle2=a3...vertically opp.angle
∆ABC~∆DBC ....BY AA
BE/EC=AE/ED
AE×CE=BE×DE
Hence Proved.

IT will help you..

Answer 2

Hello student,

Please find the answer to your question below

Given : In triangles ABC and DBC

angle A=angleD=90 degrees

AC and BD intersect each other at E

To Prove : AE x EC = ED x BE

Proof : In triangles AEB and EDC,

and AEB = DEC..............(Vertically opposite angles)

triangle ABE ~ EDC

(EB/EC) = (AE/DE)

EB x DE = EC x AE

Hence, AE x EC = ED x BE