Question

In ∆ ABC , angle ACB = 90°. Seg CD perpendicular side AB and seg CE is angle bisector of angle ACB then prove that AD/BD =AE square/BE square.

Answer 1

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,

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

Answer 2

Step-by-step explanation:

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)CDAD=BDCD=BCAC...........(1)

since CE is the bisector of ∠ACB,

By angle bisector theorem,

\begin{gathered}\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}\end{gathered}AEBE=ACBCTakingreciprocalsBEAE=BCACsquaringonbothsidesBE2AE2=BC2AC2BE2AE2=BCAC.BCACBE2AE2=CDAD.BDCD(using(1))BE2AE2=BDAD