We have a triangle $\triangle ABC$ in which angles $B$ and $C$ are greater than $45^\circ$. We construct two rectangular isosceles triangles $\triangle BAN$ and $\triangle CAM$ out of $\triangle ABC$ with angles $\angle CAM = \angle BAN = 90^\circ$. Then we draw a rectangular isosceles triangle in the triangle $ABC$ with angle $P$ equal to $90$ degrees. Prove that $\triangle MPN$ is rectangular isosceles.

My Attempt: I cannot do any thing useful because it is a very very hard question. Please answer me as soon as possible.

  • $\begingroup$ you definition of the point $P$ is not clear. Can you revise the text with more clarity? $\endgroup$
    – Chip
    May 30, 2016 at 5:18

1 Answer 1


I assume that you mean $\triangle BPC$ is a rectangular isosceles triangle with $\angle BPC=90^\circ$.

Suppose $AB=c$, $BC=a$, $CA=b$, $\angle ABC=B$, $\angle BCA=C$, $\angle CAB=A$. It is easy to see that $\angle PBN=B$, $\angle PCM=C$, $\angle MAN=180^\circ-A$.

Hence, we have $$PM^2=(a/\sqrt2)^2+(\sqrt2b)^2-2ab\cos C=b^2+c^2-(a^2/2),$$ and similarly $$PN^2=(a/\sqrt2)^2+(\sqrt2c)^2-2ca\cos B=b^2+c^2-(a^2/2).$$ Finally, $$MN^2=b^2+c^2-2bc\cos(180^\circ-A)=2(b^2+c^2)-a^2.$$

We conclude that $\triangle MPN$ is also a rectangular isosceles triangle.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .