Kazakhstan National Olympiad 2017, Final Round, 10-Grade, P4

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Kazakhstan National Olympiad 2017, Final Round, 10-Grade, P4

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Source: Kazakhstan National Olympiad 2017, Final Round, 10-Grade, P4

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#1 h Mar 18, 2017, 9:16 PM • 2 Y

Y by Adventure10, Mango247

The acute triangle $ABC$ $(AC> BC)$ is inscribed in a circle with the center at the point $O$ , and $CD$ is the diameter of this circle. The point $K$ is on the continuation of the ray $DA$ beyond the point $A$ . And the point $L$ is on the segment $BD$ $(DL> LB)$ so that $\angle OKD = \angle BAC$ , $\angle OLD = \angle ABC$ . Prove that the line $KL$ passes through the midpoint of the segment $AB$ .

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#2 h Mar 18, 2017, 10:30 PM • 2 Y

Y by Adventure10, Mango247

Denote $\measuredangle ACB=\alpha$ and $\measuredangle ABC=\beta$ , let $N=CD\cap \odot (O,OK)$ $\Longrightarrow$ $\triangle OKC\cong \triangle OND$ , by angle-chasing we get $\measuredangle NOD$ $=$ $\measuredangle BOL$ $=$ $\measuredangle COK$ $=$ $\beta-\alpha...(\star)$ $\Longrightarrow$ by $(\star)$ and $\measuredangle ODB$ $=$ $\measuredangle OBD$ $=$ $\measuredangle ONK$ $=$ $\measuredangle OKN$ $=$ $\alpha$ we get $\triangle OKN$ $\cup$ $\{D\}$ $\sim$ $\triangle ODB$ $\cup$ $\{L\}$ $\Longrightarrow \frac{DL}{LB}=\frac{DK}{DN}=\frac{DK}{KC}...(\star\star)$ If $M=KL\cap AB$ by Menelau's theorem in $\triangle BCD$ with line $\overline{KML}$ we get $DL.BM.KC=LB.MC.DK$ $\Longrightarrow$ by $(\star\star)$ we get $DL.KC=LB.DK$ hence $AM=BM$ .

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