Prove that altitude of a triangle and median of the opposite triangle belong to the same line

Question

Diagonals of some cyclic quadrilateral are perpendicular to each other and divide the quadrilateral into $4$ triangles. Prove that altitude of a triangle from the intersection of the diagonals and median of the opposite triangle also from the intersection of the diagonals belong to the same line.

Note: there is a question similar in some aspects, but I have to prove a slightly different statement.

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My attempt:

Let $ABCD$ be the given cyclic quadrilateral and let $P$ be the intersection point of the diagonals $\overline{AC}$ and $\overline{BD}$ and let's observe the opposite triangles $\Delta ABP$ and $\Delta CDP$. Let $k_1$ be the circumscribed circle of the quadrilateral $ABCD$. Then $\measuredangle PCD=\measuredangle ABP\implies\Delta ABP\sim\Delta CDP$ $\implies\measuredangle PAB=\measuredangle CDP$.

Let $T_1$ be the midpoint of the hypotenuse $\overline{AB}\iff$ $T_1$ is the center of the circumscribed circle $k_2$ of $\Delta ABP\implies$ the median $\overline{PT_1}$, as well as $\overline{AT_1}$ and $\overline{T_1B}$ is a radius of the circumscribed circle $k_2\implies\;\Delta AT_1P\;\&\;\Delta PT_1B$ are isosceles $\implies\measuredangle PAB=\measuredangle T_1PA\;\&\;\measuredangle T_1BP=\measuredangle BPT_1$.

On the other hand, let $T_2$ be the leg of the altitude of $ CDP$ from the point $P$.

$\overline{PT_2}\perp\overline{CD}\implies\measuredangle T_2PC=\measuredangle CDP\;\&\;\measuredangle DPT_2=\measuredangle PCD$.

Now we obtain: $$\color{red}{\measuredangle T_2PC}=\measuredangle CDP=\measuredangle PAB=\color{red}{\measuredangle T_1PA}$$ and $$\color{red}{\measuredangle DPT_2}=\measuredangle PCD=\measuredangle ABP=\color{red}{\measuredangle BPT_1}$$ $\measuredangle T_2PC=\measuredangle T_1PA\;\&\;\measuredangle DPT_2=\measuredangle BPT_1$ proves the statement, i.e., $T_1,P$ and $T_2$ are collinear.

Picture: May I ask if my deduction is right and how to improve my proof if necessary?

Answer 1

I think your proof works. I might summarise it with

• $\angle BPA$ is a right-angle so $\triangle BPA$ has a circumcircle with diameter $BA$ and centre $T_1$
• so $\triangle BPA$ is isosceles and $\angle T_1PA=\angle T_1AP$
• while $\angle T_1AP = \angle BAC = \angle BDC$ off chord $BC$ or the original circumcircle
• and $\angle BDC = \angle PDC = \angle T_2PC$ because of the similar triangles $\triangle PDC$ and $\triangle T_2PC$

Since $APC$ is a straight line by construction, $\angle T_1PA = \angle T_2PC$ implies $T_1PT_2$ is also a straight line.