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 Point Construction Ponder (Posted on 2015-02-25)
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..

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution | Comment 2 of 6 |
First draw a circle with diameter AB, where /ABC = 90o.

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

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