Question

two Triangles ABC and DBC are on the same base BC and on the same side of BC in which angle A is equals to angle D is equals to angle 90 degree if CA and BD meet each other at E then show that AEx EC is equals to BE x ED

Answer 1

hey buddy!

Here is your answer;

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⭐⭐⤵

Given triangles ABC and DBC are on the same base BC.

Consider, Δ’s ABC and DBC

∠A = ∠D = 90° (Given)

∠AEB = ∠DEC (Vertically opposite angles are equal)

Hence ΔABC ~ ΔDBC (AA similarity theorem)

==> AE/DE = BE/CE

∴ AE × CE = BE × DE ( Hence Proved)

➡hope this helps you deaR ✌✌✌.

Answer 2

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

Hope this will help u... :-)