Prove that triangle is isosceles, triangle that is inscribed in a circle. $AD$ is a median to side $BC$ in triangle $ABC$ that is inscribed in a circle that his center is in $(5,6)$.
Given: $ D (9,2)$ and the centroid of the triangle is $(6,5$).
There is not so much information given about the circle, thus I can't draw very good the circle (for example I don't know points in the circle, a point of the triangle or radius of the circle). There is my problem. I feel I have few given information and don't know how to start.
Can someone help me how to prove that ABC is isosceles?
 A: Sketch of the proof
Let $G$ be the centroid and $O$ the circumcentre. We know that $A$, $G$ and $D$ are collinear since they are in the median to the side $BC$.
We know that the line through $O$ and $D$ is the perpendicular bisector of the side $BC$.
Since $D$ $G$ and $O$ are collinear (the three points are in the line $y = -x + 11$), then the median to the side $BC$ and the perpendicular bisector of the side $BC$ must collide.
That means that $\angle{ADB} = \angle{ADC} = \pi/2$. One can deduce that the triangles $ADB$ and $ADC$ are congruent and so $\overline{AB} = \overline{AC}$. This completes the proof.
A: *

*Loacte O(5, 6), G(6, 5), D(9,2).





*

*Form the equation of GD. O is found to be a point on GD.

*Since ADG is a median, A must be on DG extended. From the fact that AG = 2GD. Point A is fixed.

*Draw the circum-circle (centered at O with radius OA).

*Since B and C must lie on the circum-circle of ABC, BC is a chord of that circle with D as its midpoint. Note that OD is the line from center joining the midpoint of the chord BC. This makes OD the perpendicular bisector of BC. We can therefore locate B and C.
Point 4 above also supplies enough information to conclude $\triangle ABC$ is isosceles. 
