Question

In trianlge abc , angle ACB = 90°. Seg CD perpendicular side CE and seg CE is angle bisector of angle ACB. Prove that AD/BD = AE square / BE square

Answer 1

which class this question is

Answer 2

Answer:

$\frac{{AE}^2}{{BE}^2}=\frac{AD}{BD}$

Step-by-step explanation:

Concept:

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,

$\frac{AD}{CD}=\frac{CD}{BD}=\frac{AC}{BC}...........(1)$

since CE is the bisector of ∠ACB,

by angle bisector theorem,

$\frac{BE}{AE}=\frac{BC}{AC}\\\\Taking\:reciprocals\\\\\frac{AE}{BE}=\frac{AC}{BC}\\\\squaring\:on\:both\:sides\\\\\frac{{AE}^2}{{BE}^2}=\frac{{AC}^2}{{BC}^2}\\\\\frac{{AE}^2}{{BE}^2}=\frac{AC}{BC}.\frac{AC}{BC}\\\\\frac{{AE}^2}{{BE}^2}=\frac{AD}{CD}.\frac{CD}{BD}\:\:(using(1))\\\\\frac{{AE}^2}{{BE}^2}=\frac{AD}{BD}$