Question

ABC ACB=90°, CD is perpendicular to side AB and Seg CE is angle bisector of ACB. Prove AD/BD = AE^2/ BE^2​

Answer 1

Answer:

∵ CE bisects ∠ACB

∴ From angle bisector theorem

$\frac{AE}{BE}=\frac{AC}{BC}$    ........... (1)

Again from similarity of triangles we know that ΔABC ~ ΔADC ~ ΔBDC

Therefore,

In ΔABC and ΔADC

$\frac{AB}{AC}=\frac{AC}{AD}$

or, $AC^2=AB\times AD$            ............ (2)

Similarly in ΔABC and ΔBDC we can show that

$BC^2=AB\times BD$            ............ (3)

From (2) and (3)

$\frac{AC^2}{BC^2}=\frac{AD}{BD}$

Squaring eq (1)

$\frac{AE^2}{BE^2}=\frac{AC^2}{BC^2}$

or, $\frac{AE^2}{BE^2}=\frac{AD}{BD}$  (Proved)