ABCD is cyclic iff MBCN is cyclic

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ABCD is cyclic iff MBCN is cyclic

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Source: Almaty Olympiad 2008 p2 , Kazakhstan

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#1 h Sep 28, 2018, 5:20 PM • 2 Y

Y by Adventure10, Mango247

The diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at the point $E$ , $M$ the midpoint of the segment $AE$ and $N$ the midpoint of the segment $CD$ . It is known that the diagonal $BD$ is the bisector of the angle $ABC$ . Prove that quadrilateral $ABCD$ is cyclic if and only if quadrilateral $MBCN$ is cyclic.

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#2 h Sep 28, 2018, 8:06 PM • 1 Y

Y by Adventure10

( $\Rightarrow$ ) Since $ABCD$ is cyclic, we have $\angle EAB=\angle CDB$ . Together with the equality $\angle ABE=\angle DBC$ it means that triangles $ABE$ and $DBC$ are similar. Since $BM$ , $BN$ are medians in those two triangles we obtain $\angle (BM, BN)=\angle (BA,BD)$ and since $\angle (BA,BD)=\angle (AC,CD)=\angle (MC,CN)$ we deduce that $MBCN$ is cyclic.

( $\Leftarrow$ ) From the cyclicity of $MBCN$ we get $\angle BME=\angle BNC$ . Now we observe that given an angle between the median and the side of the triangle which it bisects, and the angle at the vertex from which it emanates, the triangle is uniquely determined, up to similarity. Indeed, it is clear from the construction --- fixing length of the side bisected by the median, we may draw a median in only one way and the opposite vertex is determined uniquely as the intersection of the line containing the median and the arc such that the angle inscribed subtended by the side is as we demand. Since $\angle ABE=\angle DBC$ we deduce that $\triangle ABE\sim \triangle DBC$ and therefore $\angle BAC=\angle BAE=\angle BDC$ which means that $ABCD$ is cyclic.

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#3 h Sep 29, 2018, 1:33 PM • 1 Y

Y by Adventure10

the problem is equivalent to that we prove $BM,BN$ are isogonal wrt $\angle ABC$ which is straightforward with barycentric coordinates

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#4 h Mar 27, 2020, 4:45 AM

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If near the described quadrilateral $MBCN$ , it follows that a circle about $ABCD$ can be described, then the converse is also true. Suppose that around the quadrangles $BMCN, BCDA$ you can describe a circle. It follows that $\angle BMN = \angle BAD$ , which means that it is necessary to prove that $\Delta BMN$ and $\Delta ABD$ (2) are similar, since $BD$ is a bisector, then $\Delta BCD$ and $\Delta ABE$ are similar. Whence $\frac {BM} {BN} = \frac {AB} {BD}$ , $BM, BN$ Medians of the corresponding triangles. So the triangles (2) are really similar.

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