Question

Triangle ABC and DBC are two isosceles triangles on the same base BC. Show that $\angle ABD=\angle ACD$.

Answer 1

Hint: We first consider the theorem of equality of angles for isosceles triangle where angles opposite to the equal sides of an isosceles triangle are also equal. We consider the $\Delta ABC$ and $\Delta DBC$ to find their equal angles. Then we add them to prove the given relation.

Complete step by step answer:
$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base BC.
We know that the angles opposite to the equal sides of an isosceles triangle is also equal.
For $\Delta ABC$, the sides AB and AC are equal. The opposite angles of those sides are $\angle ABC$ and $\angle ACB$ respectively. Therefore, $\angle ABC=\angle ACB$.
For $\Delta DBC$, the sides DB and DC are equal. The opposite angles of those sides are $\angle DBC$ and $\angle DCB$ respectively. Therefore, $\angle DBC=\angle DCB$.
We got two equations of equality of angles.
We add them and get $\angle ABC+\angle DBC=\angle ACB+\angle DCB$.
Looking at the image we can say that the angles $\angle ABD$ and $\angle ACD$ can be broken into two parts.
We have $\angle ABD=\angle ABC+\angle DBC$ and $\angle ACD=\angle ACB+\angle DCB$.
We know that $\angle ABC+\angle DBC=\angle ACB+\angle DCB$ which gives that $\angle ABD=\angle ACD$.
Thus proved $\angle ABD=\angle ACD$.

Note: We need to be careful about adding wrong angles with one another. In addition if we had added $\angle ABC+\angle DCB$, then we wouldn’t have gotten neither $\angle ABD$ nor $\angle ACD$. Similar things can be said for $\angle DBC+\angle ACB$.