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# $$\triangle{ABC}$$ is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that $$\angle{BCD}$$ is a right angle. In $$\triangle{ABC}$$, we have,

AB = AC ...(i)(Given)
$$\angle{ACB}$$ = $$\angle{ABC}$$ ...(ii)($$\because$$ angles opposite to equal sides are equal)

From (i) and (iii), AC = AD

Now, in $$\triangle{ADC}$$ , we have,

$$\angle{ACD}$$ = $$\angle{ADC}$$ ...($$\because$$ angles opposite to equal sides are equal)

Also, $$\angle{ACD}$$ = $$\angle{BDC}$$ ...(iv)($$\because \angle{ADC} = \angle{BDC}$$)

On adding Equations (ii) and (iv), we get,

$$\Rightarrow \angle{ACB} + \angle{ACD} = \angle{ABC} + \angle{BDC}$$
$$\Rightarrow$$ $$\angle{BCD}$$ = $$\angle{ABC}$$ + $$\angle{BDC}$$

Adding $$\angle{BCD}$$ on both sides, we have,

$$\Rightarrow \angle{BCD} + \angle{BCD} = \angle{ABC} + \angle{BDC} + \angle{BCD}$$
$$\Rightarrow$$ 2 $$\angle{BCD}$$ = $$180^\circ$$ ...($$\because$$ sum of all angles of a triangle is $$180^\circ$$)

Therefore, $$\angle{BCD}$$ = $$90^\circ$$
Hence, proved.