$ABCD$ is a cyclic quadrilateral. The midpoints of the diagonals $AC$ and $BD$ are respectively $P$ and $Q$. If $BD$ bisects $\angle AQC$, the prove that $AC$ will bisect $\angle BPD$
Source- NMTC Junior 2023 India
5
-
1$\begingroup$ Your title mentions the "harmonic quadrilateral" but the question itself does not mention it. Can you clarify what you are asking exactly? $\endgroup$Mikhail Katz– Mikhail Katz2026-02-19 16:36:23 +00:00Commented Feb 19 at 16:36
-
$\begingroup$ This condition works in harmonic quadrilaterals. If I were to state only cyclic quadrilateral, the question would be wrong. The question is solved by proving the quadrilateral harmonic. I had a doubt earlier regarding the construction of tangents to prove $ABCD$ is harmonic. If my solution is correct, I will close this question. $\endgroup$JKH_Mathematics– JKH_Mathematics2026-02-19 16:43:54 +00:00Commented Feb 19 at 16:43
-
$\begingroup$ I see you've added your attempt as an answer. Why don't you add it to your question instead? $\endgroup$Andrei– Andrei2026-02-19 16:56:16 +00:00Commented Feb 19 at 16:56
-
1$\begingroup$ I didn't come up with it when I posted the question that's why. I would love to see some other solution although projective solution is available in AOPS. $\endgroup$JKH_Mathematics– JKH_Mathematics2026-02-19 17:01:12 +00:00Commented Feb 19 at 17:01
-
1$\begingroup$ By the way, it is OK to self-answer your own question. In particular, if the solution was found only after posting the question, you can indeed post it as an answer. math.stackexchange.com/help/self-answer "Stack Exchange has always explicitly encouraged users to answer their own questions. If you have a question that you already know the answer to, and you would like to document that knowledge in public so that others (including yourself) can find it later, it's perfectly okay to ask and answer your own question on Mathematics Stack Exchange." $\endgroup$mezzoctane– mezzoctane2026-02-19 18:44:22 +00:00Commented Feb 19 at 18:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.
Observe that the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.
Claim: $AG,BD,CG$ are concurrent at $G$.
Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ and iff $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.
Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.
$\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.