Question

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 AE×EC=BE×BD

Answer 1

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.