Question

in traingle abc angleacb=90 cd perpendicular ab ce is angle bisector of angle acb prove that ad/bd=ae2/be2​

Answer 1

Question:

Please see the attached diagram

CD is perpendicular to AB.

∠ACE = ∠BCE, CE is angular bisector of angle C.  

To prove that AD/BD = AE²/BE²

Step-by-step explanation:

On ΔABC and ΔACD  

∠ACB = ∠ADC,  

∠DAC = ∠CAB  

So ΔABC is similar to ΔACD  

We can derive from similarity of triangle rule,  

AC/AB=AD/AC  

AC²= AB×AD --------------------------E1  

similarly ΔBCD and ΔBAC are also similar triangles.  

We can derive from similarity of triangle rule,  

BC/BA = BD/BC  

BC²= BA×BD -----------------------------E2  

∴ Dividing E2 by E1 we get  

BC²/AC²=AB×BD/AB×AD  

∴ BC²/AC² =BD/AD -----------------------------------E3

In ΔACB, CE is the angle bisector of ∠ACB.

Hence from Triangle bisector theorem, E cuts AB in the ratio of sides of triangle.

That is AE/BE = AC/BC.

Squaring we get, BC²/AC² = AE²/BE² ---------------------------E4

From E3 and E4, we can say

AD/BD = AE²/BE²

Hence proved.  

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,

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