Question

in triangle abc, angle acb = 90°. sec CD is perpendicular to side AB and seg CE is angle bisector of angle ACB Prove that AD/ BD = AE×AE/ BE×BE ​

Answer 1

only interchange the letters and also let me know if you have got it right

Answer 2

Solution:-

Angle bisector theorem:

When vertical angle of a triangle is bisected, the bisector divides the base into two segments which have the ratio as the order of other two sides.

From diagram, it is clear that

ΔADC and ΔCDB are similar.

Then,

AD/ CD = CD/BD = AC/BC -----(I)

since CE is the bisector of ∠ACB,

by angle bisector theorem,

BE/ AE = BC/AC

taking reciprocal

AE/BE = AC/BC

Both side square

AE^2/ BE^2 = AC^2/ BC^2

AE^2/ BE^2 = AC/ BC× AC/ BC

AE^2/ BE^2 = AD/ CD × CD/ BD

AE^2 \ BE^2 = AD / BD

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@GauravSaxena01