prove that the intersection of the two lines $l_1$ and $l_2$ is concyclic with midpoints of three sides of a triangle

Let $ABC$ be a triangle. Suppose the lines $l_1$ and $l_2$ are perpendicular, and meet each side (or its extension) in a pair of points symmetric across the midpoint of the side. Then the intersection of $l_1$ and $l_2$ is concyclic with the midpoints of the three sides.

Let $M_A, M_B,M_C$ be the midpoints of the sides $BC,CA,AB$ and let $P= l_1\cap l_2$. I'm not really sure how to visualize this geometrically, so could someone draw a good picture to represent the situation? I think that could clear up a lot of confusion I've been having. In a solution I found online, it said the lines $l_1, l_2, BC$ should form a right triangle where $M_B$ is the midpoint of the hypotenuse and the triangle formed by the points $P, M_B, l_2\cap BC$ is isoceles with $\angle (M_BP, l_2) = \angle (l_2, BC)$. Similarly, $\angle (l_2, M_AP) = \angle (CA,l_2)$. Then one could add these inequalities to show $\angle M_B P M_A + \angle M_BM_CM_A = 180$ so that the quadrilateral $M_B PM_AM_C$ is cyclic.

But again, I'm not sure whether these claims hold.

Solution 1:

A good diagram surely helps. First draw $\ell_1$ cutting sides $AB,BC,CA$ of $\triangle ABC$ in $X,Y,Z$ respectively. Now take points $X',Y'$ on $AB,BC$ respectively so that $XM_C=X'M_C$ and $YM_A=Y'M_A$. Then the condition that $\ell_1,\ell_2$ meet the sides symmetrically across midpoints mean, $\ell_2$ pass through $X',Y'$.

It should be considered an exercise, preliminary to current question, to show that intersection of $\ell_2$ with $AC$, $Z'$, lies symmetrically to $Z$, that is, $ZM_B=Z'M_B$.

For such a configuration, the angle between $\ell_1,\ell_2$ could vary; here it is given that $\angle(\ell_1,\ell_2)=90^\circ$. For this constraint, one has to show that their intersection point $P$ lies on the circle through the three midpoints.

It is clear that $\angle M_AM_CM_B = \angle C$. For right $\triangle ZPZ'$, $M_B$ is circumcenter and $PM_B=Z'M_B$. So $\angle M_BPZ'= \angle M_BZ'P$. Similarly $PM_AY'$ is isosceles and $\angle M_APY' = \angle M_AY'P=\angle CY'Z'$.

In $\triangle Y'CZ'$, we see the exterior angle $$\angle C = \angle CY'Z' + \angle CZ'Y' =\angle M_APY' + \angle M_BPZ' = \angle M_APM_B $$

Thus $\angle M_APM_B = \angle M_AM_CM_B$ and it follows that $P$ lies on the said circle.