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Point Construction Ponder (Posted on 20150225) 

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..
Solution

 Comment 2 of 6 

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 20150225 16:09:56 


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