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.