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# ABC and DBC are isosceles triangles on the same base BC (see figure). Show that $$\angle{ABD}$$ = $$\angle{ACD}$$.

In $$\triangle{ABC}$$, we have,

AB = AC ...($$\because$$ $$\triangle{ABC}$$ is an isosceles triangle)

$$\therefore$$ $$\angle{ABC}$$ = $$\angle{ACB}$$ ...(i)($$\because$$ angles opposite to equal sides are equal)

Similarly, in $$\triangle{DBC}$$, we have,

BD = CD ...($$\because$$ $$\triangle{DBC}$$ too, is an isosceles triangle)

$$\therefore$$ $$\angle{DBC}$$ = $$\angle{DCB}$$ ...(ii) ($$\because$$ angles opposite to equal sides are equal)
Now, On adding, Equations (i) and (ii), we get,

$$\Rightarrow \angle{ABC} + \angle{DBC} = \angle{ACB} + \angle{DCB}$$
$$\therefore$$ $$\angle{ABD}$$ = $$\angle{ACD}$$
Hence, proved.