Question

In ∆ABC,the perpendicular drawn from C on the hypotenuse,meet the hypotenuse at D. Bisector of angle C meets the hypotenuse at E. Prove that
AD/DB=(AE)^2/(EB)^2

Answer 1

Proved that,  $\frac{AE^2}{BE^2} = \frac{AD}{BD}$.

Given

To prove that , $\frac{AD}{DB} = \frac{AE^{2} }{EB^{2}}$ .

From the figure,

                         Segment CE and CD bisects ∠ACB.

In ΔACB,

               If segment CE bisects ∠ACB.

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

               Squaring on both the sides,

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

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

                 $\frac{AE^2}{BE^2} = \frac{AC^2}{BC^2}$ -----> (1)

In ΔACB,

               If segment CD bisects ∠ACB.

                 m∠ACB = 90°

Where, ΔABC ≅ ΔADC ≅ ΔBDC

            In  ΔABC and ΔADC

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

              $AC^2 =$ AB × AD -----> (2)

            In ΔABC and ΔBDC

               $\frac{BC}{AB} = \frac{BD}{BC}$

               $BC^2$ = BD × AB -----> (3)

From equation (1),

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

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

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

Hence, proved that  $\frac{AE^2}{BE^2} = \frac{AD}{BD}$ .

To learn more...

1. brainly.in/question/2813200

2. brainly.in/question/8757272