Answer to Question #100384 in Geometry for Mark

Question

Triangle ABC is inscribed in a circle. A median AM is drawn so that it intersects
the circle at point D. It is known that AB=1 and BD=1. Find BC.

Answer 1

1.Since the triangle is inscribed in a circle, its median is the midpoint perpendicular. $∠AOB = 90°$

2. As $AO$ is a common side of $ΔAOC$ and $ΔAOB$, $CO=OB$ and $∠AOC=∠A0B=90°$, so $AC=AB=1$.

3 As we see, $ACDB$ is inscribed in a circle and it's diagonals intersect at right angles, all sides $(AC,CD,DB,AB)$ are the same, so $ABCD$ is a square whose diagonal is equal $\sqrt{2}$

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