Consider a right triangle ABC and construct a point P inside the triangle such that the angles PBC, PCA and PAB are equal.

*** Adapted from a problem which appeared in a Hungarian Math Contest..

First draw a circle with diameter AB, where /ABC = 90^o.

P must lie on this circle since then /PBC = /PAB (angle

between tangent BC and chord BP is equal to angle in

the alternate segment).

Then draw a circle through B and C with AC as its

tangent. Its centre lies at the intersection of the

perpendicular bisector of BC and the line through C

perpendicular to AC. P must also lie on this circle since

then  /PCA = /PBC   (angle in alternate segment).

So P is the point of intersection of the two circles.

Posted by Harry on 2015-02-25 16:09:56