# ABC is an isosceles triangle -prove AD=BC

ABC is an isosceles triangle having $\angle B=\angle C=2*\angle A$. If BD bisecting $\angle B$ meets AC in D,prove that AD=BC.

I know congruent triangles would help but am not able to figure out how to use them. ADB can not be congruent to CBD. I am trying to figure out which triangles might be congruent(triangles with AD as side and BC as side).

See angles are $72,72,36$ of triangle ABC . So BD is bisector of B implies each angle=36. So $AD=BD ...(1)$isoceles triangle theorem now in triangle BDC angle $BDC=72$ but also $DCB =72$ so $BD=BC$..(2) isoceles triangle theorem thus from 1,2 $AD=BC$ thas all.

Instead of looking for congruent triangles, you should instead find the angle of triangles themselves.

We know that : $$2\times \angle A=\angle B =\angle C$$

Since $\angle A, \angle B , \angle C$ are parts of single triangle then : $$\angle A+\angle B +\angle C=\pi$$ solve for these angles we would find: $$\angle A= \frac{\pi}{5}, \angle B =\angle C=\frac{2\pi}{5}$$ $BD$ bisect $\angle B$ means that

$$\angle CBD = \angle ABD =\frac{\pi}{5}$$ Which tells us that triangle $ABD$ is isosceles thus $$AD =BD$$

We can also solve for $\angle BDC$ to find that $$\angle BDC=\frac{2\pi}{5}$$ thus triangle $BDC$ is also isosceles and $$BD=BC$$ finally we proved that $$AD=BC$$

By construction we have:

$$\angle DBA=\angle DCB=\frac{1}{2} \angle ABC=\frac{1}{2} \angle ACB=\angle CAB$$

so, for the triangle $ADB$: $$\angle DAB= \angle CAB=\frac{1}{2} \angle ABC=\angle DBA \Rightarrow DA=DB$$ and for the triangle $CDB$: $$\angle CDB=180°-(\angle DCB+\angle DBC)=(180°-\angle DBC)-\angle DCB=$$ $$=(180°-\angle CAB)-\angle DCB=2\times \angle DCB-\angle DCB=\angle DCB \Rightarrow DB=CB$$

So: $DA=DB$