In a triangle $ABC$ with side $AB=AC$ and $\measuredangle BAC=20 ^\circ $. $D $ is a point on side $AC$ and $BC = AD$. Find $ \measuredangle DBC$

Question

Problem : In a triangle $ABC$ with side $AB=AC$ and $\measuredangle BAC=20 ^\circ $.

$D $ is a point on side $AC$ and $BC = AD$. Find $ \measuredangle DBC$

Solution:

$AB =AC$

So $ \measuredangle ACB = \measuredangle ABC$

$ \measuredangle ACB = \measuredangle ABC=80^\circ$ (Using angle sum property)

I am unable to continue from here.

Any assistance is appreciated.

Answer 1

Let $E$ be a point inside the triangle such that $EB=EC=BC$.

$\qquad\qquad\qquad$

Since $\measuredangle{ABE}=\frac{180^\circ-20^\circ}{2}-60^\circ=20^\circ$, we can see that $\triangle{ABD}$ and $\triangle{BAE}$ are congruent.

Hence, $$\measuredangle{DBC}=\measuredangle{ABC}-\measuredangle{ABD}=\measuredangle{ABC}-\measuredangle{BAE}=80^\circ-10^\circ=\color{red}{70^\circ}.$$