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Let an acute triangle ABC be given. Prove that the circles whose diameters are AB and AC have a point of intersection on BC. How do I go about this problem? Can You Please Give Me a Hint?

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Does this only apply for acute triangles? I think it applies for all. – Ronald Mar 31 '12 at 23:51

Hint: Angles in a semi-circle are $90^{\circ}$.

(more precisely, the locus of $P$ such that $\angle{XPY}$ is $90^{\circ}$ is a circle with diameter $XY$).


Drop a perpendicular from A to BC, let the foot of the perpendicular be D. Use the above and see that D lies on both the circles!

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One approach is to construct the triangle that joins the midpoints of AB, AC, and BC.

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Somewhat related to Ross's answer, you know that one point of intersection of the circles is point $A$. The line segment joining the two points of intersection is a chord of both circles and is perpendicularly bisected by the segment joining the centers of the two circles. How does the segment joining the two centers of the circles relate to $\overline{BC}$?

The line segment joining the two centers of the circles is parallel to $\overline{BC}$. Combining this with the fact that that segment is the perpendicular bisector of the chord joining the points of intersection gives the desired result.

Also, this same reasoning works in the case where $\angle B$ or $\angle C$ is not acute (in which case the point of intersection isn't on $\overline{BC}$, but is still on $\overleftrightarrow{BC}$.

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