ABCD with angles ABC and BCD equal to 120

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ABCD with angles ABC and BCD equal to 120

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Source: Ukraine MO 2015 grade 11 P4

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#1 h Jan 11, 2018, 1:36 PM • 2 Y

Y by Adventure10, Mango247

In convex quadrilateral $ABCD$ with angles $ABC$ and $BCD$ equal to $120^{\circ}$ , $O$ is intersection of diagonals, $M$ is midpoint of $BC$ . $K$ is intersection of segments $MO$ and $AD$ . It's known that $\angle BKC=60^{\circ}$ . Prove that $\angle BKA=\angle CKD = 60^{\circ}$ .

Proposed by Nazar Serdyuk

This post has been edited 1 time. Last edited by richrow12, Jan 12, 2018, 4:18 PM

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#2 h Jan 12, 2018, 1:12 PM • 1 Y

Y by Adventure10

Anyone?
Is this geometry hard?

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#3 h Jan 12, 2018, 2:20 PM • 3 Y

Y by richrow12, Adventure10, Mango247

I think you actually mean $\angle BKD=\angle CKA= 120^{\circ}$ .

Now, let $P$ and $Q$ be on lines $\overline{BA}$ and $\overline{CD}$ so that $\angle BKP=\angle CKQ = 60^{\circ}$ , it's easy to see that triangles $KBP$ , $KCB$ ; $KQC$ are similar, let $R$ the intersection of lines $\overline{BQ}$ and $\overline{CP}$ , because of the similarity of $KBP$ and $KQC$ , we get that quadrilaterals $PBRK$ and $KRCQ$ are cyclic, so, $\angle RBK=\angle RPK$ which implies that $\angle CBR=\angle BAR$ , this means that $BC$ is tangent to $PBR$ and , analogously, is also tangent to $CQR$ , and so, $K$ , $R$ ; $M$ and $O$ are collineal, and finally it's clear to see that $R$ and $O$ must be the same points (otherwise $P$ , $K$ and $Q$ would not be collinear) and finally $\angle BKD=\angle CKA= 120^{\circ}$ .

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#4 h Jan 12, 2018, 4:18 PM • 2 Y

Y by Adventure10, Mango247

Thanks, edited.

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