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ΔABC and ΔDBC are two triangles on same base BC and on the same side of BC with angle A = angle D = 90°. If CA and BD meet each other at E, show that



Consider ΔAEB and ΔDEC
\angle A = \angle D = 90^o

\angle AEB = \angle DEC [vertically opp angles]

Therefore, ΔAEB ~ ΔDEC (by AA similarity)

\Rightarrow  \frac{AE}{DE} =  \frac{EB}{EC}

On cross multiplication, 

AE × EC = DE × EB

Note: As per the question AE ×EC = BE × BD, which is , I think, not posible.
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