Question

9. In Fig. 6.63, D is a point on side BC of AABC
Prove that AD is the
BD AB
such that
CD AC
bisector of BAC.​

Answer 1

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Ex 6.6,9

In Fig. 6.63, D is a point on side BC of ΔABC such that$\frac{BD}{CD} = \frac{AB}{AC}$. Prove that AD is the bisector of ∠BAC.

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Const. Draw AE=AC , join EC

in ΔAEC

AE=AC {Const.}

∴ ∠AEC=∠ACE {Isosceles Triangle Prop.}

——————(1)

$\frac{BD}{CD} = \frac{AB}{AC}$———Given

putting AC=AE {Const.}

$\frac{BD}{CD} = \frac{AB}{AE}$

In ΔBEC

$\frac{BD}{CD} = \frac{AB}{AE}$——— proved above

∴ AD||EC

so,

∠ACE = ∠CAD———(2)

∠AEC = ∠BAD———(3)

from (1) , (2) & (3)

∠CAD = ∠BAD

thus,

AD is the bisector of ∠BAC.

Hence Proved.

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