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

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.

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?

• $$\angle BPA$$ 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.