# Proving that a triangle is isosceles.

In a triangle $$ABC$$, let $$D$$ be a point on the segment $$BC$$ such that $$AB + BD = AC + CD$$. Suppose that the points $$B,C$$ and the centroids of triangles $$ABD$$ and $$ACD$$ lie on a circle. Prove that $$AB = AC$$.

I began this way: Denote the centroid of triangle $$ABD$$ by $$G_1$$ and the centroid of triangle $$ACD$$ by $$G_2$$. Then by using $$Basic$$ $$Proportionality$$ $$Theorem$$, I got that $$BG_1G_2C$$ is an isosceles trapezium. After that I am stuck. Please help.

• Try to prove that BD=CD – Moti Jan 1 '19 at 7:08
• Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation) – Yellow Jan 1 '19 at 7:39

Now, in the standard notation $$c+BD=b+a-BD,$$ which gives $$BD=\frac{a+b-c}{2}$$ and $$CD=\frac{a+c-b}{2}.$$
Thus, $$CG_2=\frac{1}{3}\sqrt{2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2}$$ and $$BG_1=\frac{1}{3}\sqrt{2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2},$$ which gives $$2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2=2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2$$ or $$(b-c)(b+c-a)=0$$ or $$b=c.$$