In the circumscircle of a triangle $ABC$, let $A_1$ be the point diametrically opposed to the vertex $A$. Let $A'$ the intersection point of $AA'$ and $BC$. The perpendicular to the line $AA'$ from $A'$ meets the sides $AB$ and $AC$ at $M$ and $N$, respectively. Prove that the points $A,M,A_1$ and $N$ lie on a circle which has the center on the height from $A$ of the triangle $ABC$. This was the statement

My solution:

First we prove that $ANA_1M$ lie on the same circle. Observe that $A'NCA_1$ are concyclic, since $\angle NA'A_1=\angle NCA_1=90^o$, from here $\angle NA_1A'=\angle ACB$. Now it is easy to see that $\angle ANA'=\angle ABC$, hence $\angle AMN=180-(\angle BAC+\angle ABC)=\angle ACB=\angle NA_1A'$ so they lie in the same circle.

Now, in order to proof that the center of the circle lie on the height from $A$ of the triangle $ABC$ we draw it, and call $P$ the intersection of it to the circle $AMA_1N$. Lets evaluate the angle $\angle AA_1P$. Since $PANA_1$ li on the same circle $\angle NA_1P=180-\angle NAP=180-(90-\angle ACB)=90+\angle ACB$. Now $\angle AA_1P=\angle NA_1P-\angle NA_1A'=90+\angle ACB-\angle ACB=90^o$ so $APA_1$ is a right-angle triangle whose circumcircle is the same as $AMA_1N$ hence the center lies on $AP$ which we said to be the height from $A$. An pretty easy problem.

Is my proof correct?

• Yes, your proof is totally correct. – Amir Naseri Apr 1 '16 at 17:08
• @AmirNaseri Thanks! – Weijie Chen Apr 1 '16 at 17:09
• Besides a typo, seems to be OK. However, the first part can be slightly shortened. After proving $A_1, A’, N, C$ are con-cyclic, $A, M, A_1, N$ are also con-cyclic because $\angle MAA_1 = \angle BAA_1 = \angle A’CA_1 = \angle A’NA_1$. – Mick Apr 1 '16 at 18:25
• @Mick But after all its OK rght? – Weijie Chen Apr 1 '16 at 18:27
• Yes and is a good job. – Mick Apr 2 '16 at 16:20