$ \triangle \mathrm{ABC} $ is an isosceles triangle in which $ \mathrm{AB}=\mathrm{AC} $. Side $ \mathrm{BA} $ is produced to $ \mathrm{D} $ such that $ \mathrm{AD}=\mathrm{AB} $ (see Fig. 7.34). Show that $ \angle \mathrm{BCD} $ is a right angle. "
Given:
$\triangle ABC$ is an isosceles triangle in which $AB=AC$. Side $BA$ is produced to $D$ such that $AD=AB$.
To do:
We have to show that $\angle BCD$ is a right angle.
Solution: Let us consider $\triangle ABC$,
Given,
AB = AC
We know that,
The angles opposite to the equal sides are also equal.
This implies,
$\angle ACB = \angle ABC$
Now, let us consider $\triangle ACD$
Given,
AD = AB
We know that,
The angles opposite to the equal sides are also equal.
This implies,
$\angle ADC= \angle ACD$
We know that,
The sum of the interior angles of a triangle is always equal to $180^o$
This implies
In $\triangle ABC,$
$\angle CAB+\angle ACB+\angle ABC = 180^o$
Therefore,
$\angle CAB + 2\angle ACB=180^o$
This implies,
$\angle CAB = 180^o–2\angle ACB$.........(i)
Similarly,
In $\triangle ADC$,
$\angle CAD = 180^o – 2\angle ACD$......(ii)
also, as $BD$ is a straight line,
We get,
$\angle CAB + \angle CAD = 180^o$
By adding (i) and (ii) we get,
$\angle CAB+\angle CAD=180^o–2\angle ACB+180^o–2\angle ACD$
$180^o=360^o–2\angle ACB-2\angle ACD$
$2(\angle ACB+\angle ACD)=180^o$
$\angle BCD=90^o$.
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