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: enter image description here May I ask if my deduction is right and how to improve my proof if necessary?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.