In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$. If $\angle B = 60^o$ and $\angle A = 70^o$, prove that $AD > CD$.

Given:

In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$.

$\angle B = 60^o$ and $\angle A = 70^o$.

To do:

We have to prove that $AD > CD$.

Solution:

In $\triangle ABC$,

$\angle CBD + \angle CBA = 180^o$                   (Linear pair)

$\angle CBD + 60^o = 180^o$

$\angle CBD = 180^o - 60^o = 120^o$

In $\triangle BCD$,

$BD = BC$

$\angle CDB = \angle BCD$             (Angles opposite to equal sides are equal)

$\angle CDB + \angle BCD = 180^o - 120^o = 60^o$

$\angle CDB = \angle BCD = \frac{60^o}{2} = 30^o$

In $\triangle ABC$,

$\angle A + \angle B + \angle C = 180^o$

$70^o + 60^o + \angle C = 180^o$

$\angle C = 180^o-130^o$

$\angle C = 50^o$

This implies,

$\angle ACD = \angle ACB + \angle BCD$

$ = 50^o + 30^o$

$= 80^o$

In $\triangle ACB$,

$\angle ACD = 80^o$ and $\angle A = 70^o$

Side opposite to greatest angle is the largest side)

Therefore, $AD > CD$

Hence proved.

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Updated on: 10-Oct-2022

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