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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.