A Segment Bisection Problem

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A Segment Bisection Problem

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Source: VN Math Olympiad For High School Students P9 - 2025

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#1 h Apr 16, 2025, 1:36 AM

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In triangle $ABC$ , let the incircle $\omega$ touch sides $BC, CA, AB$ at $D, E, F$ , respectively. Let $P$ lie on the line through $D$ perpendicular to $BC$ . Let $Q, R$ be the intersections of $PC, PB$ with $EF$ , respectively. Let $K, L$ be the projections of $R, Q$ onto line $BC$ . Let $M, N$ be the second intersections of $DQ, DR$ with the incircle $\omega$ . Let $S$ be the intersection of $KM$ and $LN$ . Prove that the line $DS$ bisects segment $QR$ .

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#2 h Apr 16, 2025, 7:03 AM • 1 Y

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buratinogigle wrote:

In triangle $ABC$ , let the incircle $\omega$ touch sides $BC, CA, AB$ at $D, E, F$ , respectively. Let $P$ lie on the line through $D$ perpendicular to $BC$ . Let $Q, R$ be the intersections of $PC, PB$ with $EF$ , respectively. Let $K, L$ be the projections of $R, Q$ onto line $BC$ . Let $M, N$ be the second intersections of $DQ, DR$ with the incircle $\omega$ . Let $S$ be the intersection of $KM$ and $LN$ . Prove that the line $DS$ bisects segment $QR$ .

This is my solution during the test

Let $EF$ cut $BC$ at $T$ . Note that $(BC, DT) = -1$ , then $D(TP, QR) = P(TD, QR) = P(TD, CB) = -1$ , therefore $DT$ is the external bisector of angle $RDQ$ , which also implies that $DM = DN$ , so we get $MN \parallel BC$ .
We have $D(TS, QR) = D(TS, MN) = \dfrac{DL}{DK} = \dfrac{DQ}{DR} = \dfrac{TQ}{TR}$ .
Therefore, $DS$ bisects $QR$ .
This problem will work with any $M$ and $N$ lie on $DQ$ , $DR$ satisfy $MN \parallel BC$ .

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#4 h Apr 16, 2025, 11:54 AM

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Why is $DT$ the external bisector?

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#5 h Apr 16, 2025, 1:54 PM

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用调和，然后表达一下比例

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#6 h Today at 4:53 AM

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Here is the official solution of mine.

Disregard the simple case when $EF \parallel BC$ . Assume that $EF$ intersects $BC$ at $G$ . Applying Menelaus’ Theorem to triangle $PBC$ with collinear points $G, Q, R$ , we have
$\frac{GB}{GC} \cdot \frac{QC}{QP} \cdot \frac{RP}{RB} = 1,$ which is equivalent (as a consequence of Thales' Theorem) to
$\frac{DB}{DC} \cdot \frac{LC}{LD} \cdot \frac{KD}{KB} = 1,$ or
$\frac{LD}{KD} = \frac{DB}{KB} \cdot \frac{LC}{DC} = \frac{DP}{RK} \cdot \frac{QL}{DP} = \frac{QL}{RK}.$ From this, the two right triangles $\triangle DKR$ and $\triangle DLQ$ are similar. As a consequence, $\angle RDK = \angle QDL$ , which implies $MN \parallel KL$ . Let $T$ be the midpoint of $QR$ . From the similarity of triangles $DKR$ and $DLQ$ , and by applying the trigonometric form of Ceva's Theorem, we obtain
$\frac{\sin\angle QDT}{\sin\angle RDT} \cdot \frac{\sin\angle LND}{\sin\angle LNM} \cdot \frac{\sin\angle KMN}{\sin\angle KMD} = \frac{DR}{DQ} \cdot \frac{\sin\angle LND}{\sin\angle NLD} \cdot \frac{\sin\angle MKD}{\sin\angle KMD} = \frac{DR}{DQ} \cdot \frac{DL}{DN} \cdot \frac{DM}{DK} = 1.$ Thus, the lines $DT$ , $KM$ , and $LN$ are concurrent.

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