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$AD$ 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?

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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.

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  • $\begingroup$ I was sure that one perpendicular bisector and median must coincide... but how do I know it's that one? $\endgroup$ – Pichi Wuana Jan 8 '17 at 15:19
  • $\begingroup$ @PichiWuana Perhaps this helps a bit more. $\endgroup$ – Darth Geek Jan 8 '17 at 15:27
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  1. Loacte O(5, 6), G(6, 5), D(9,2).

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

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

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

  4. 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.

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