Track / Stage 7 / 183 of 300 #1583 of 1964

Problem 1583

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.3 Prove it

The incircle of a triangle $ABC$ is tangent to its sides $AB,BC,CA$ at $M,N,K,$ respectively. A line $l$ through the midpoint $D$ of $AC$ is parallel to $MN$ and intersects the lines $BC$ and $BD$ at $T$ and $S$ , respectively. Prove that $TC\equal{}KD\equal{}AS.$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

1. Identify Key Points and Relationships:
- Let $I$ be the incenter of $\triangle ABC$ .
- The incircle is tangent to $AB$ at $M$ , to $BC$ at $N$ , and to $CA$ at $K$ .
- $D$ is the midpoint of $AC$ .
- Line $l$ through $D$ is parallel to $MN$ and intersects $BC$ at $T$ and $BD$ at $S$ .

2. Use Thales' Theorem:
- Since $l \parallel MN$ , by Thales' theorem, the segments $DT$ and $DS$ are proportional to the segments $MN$ and $MK$ .

3. **Prove $TC = KD = AS$ :**
- Since $D$ is the midpoint of $AC$ , $AD = DC$ .
- Because $l \parallel MN$ , the triangles $\triangle DTC$ and $\triangle DNM$ are similar.
- Therefore, $\frac{DT}{DN} = \frac{DC}{DM}$ .
- Since $D$ is the midpoint, $DC = \frac{AC}{2}$ .
- Similarly, $\triangle DAS$ and $\triangle DMK$ are similar, giving $\frac{DS}{DK} = \frac{DA}{DM}$ .

4. Calculate Lengths:
- Since $D$ is the midpoint, $DA = DC = \frac{AC}{2}$ .
- By similarity, $DT = DC$ and $DS = DA$ .

5. Conclude the Equalities:
- Therefore, $TC = KD = AS$ .

$\boxed{TC = KD = AS}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.