angle chasing in RMO, cyclic ABCD, 2 circumcircles, incenter, right wanted

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angle chasing in RMO, cyclic ABCD, 2 circumcircles, incenter, right wanted

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Source: CRMO 2015 region 1 p1

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#1 h Sep 30, 2018, 1:18 AM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, Rounak_iitr

In a cyclic quadrilateral $ABCD$ , let the diagonals $AC$ and $BD$ intersect at $X$ . Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$ . If $X$ is the incentre of triangle $ABY$ , show that $\angle CAD = 90^o$ .

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Synthetic_Potato

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Synthetic_Potato

#2 h Sep 30, 2018, 3:29 AM • 4 Y

Y by nguyendangkhoa17112003, AlastorMoody, SK1729, Adventure10

See that $\angle DCX = \angle ABX=\angle XBY = \angle XCY$ Hence, $Y$ lies on $CD$ . Now $\angle XYD = \angle XBC = \angle XAD = \angle XYC$ . Since $Y \in CD$ , we conclude that $\angle XAD=90^\circ$ . Hence $\angle CAD=90^\circ$ . $\blacksquare$ .

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#3 h Sep 30, 2018, 6:18 AM • 2 Y

Y by Adventure10, Mango247

Claim.1 : $Y \in CD$ (like Synthetic_Potato):
We have: $\angle{YCX}=\angle{YBX} =\angle{XBA} =\angle{DCX}$ $\Longrightarrow Y \in CD$ Claim.2 : $\angle{DYX} =90$ :
Let $K=AB \cap DC$ .
Notice that $Y$ is the miquel point of the cyclic quadrilateral $ACBD$ . $\Longrightarrow \angle{KYX} =90 \Longrightarrow \angle{DYX} =90 \Longrightarrow \angle{CAD} =90$ .
As desired...
$\blacksquare$

This post has been edited 2 times. Last edited by Anaskudsi, Sep 30, 2018, 6:18 AM

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Pigeonhole_Biswaranjan_

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Pigeonhole_Biswaranjan_

#4 h Sep 30, 2018, 6:34 AM • 2 Y

Y by Adventure10, Mango247

$\angle YAB=2m$ for some m,Notice, $\angle YXB= 90+ .5( \angle YAB)=90+m.$ In triangle YAB .
$\angle YXD=180- \angle YXB=90-m$
.As ABCD is cyclic, $\angle CAB=\angle CDB=m.$
In triangle YDX $,\angle CYX=\angle CDX +\angle DXY= m+ 90-m=90.$
But, $\angle CYX =\angle CAD(cyclic quad.)$
hence proved

This post has been edited 5 times. Last edited by Pigeonhole_Biswaranjan_, Sep 30, 2018, 6:47 AM

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#5 h Apr 18, 2019, 12:41 PM • 2 Y

Y by guptaamitu1, Adventure10

Join $XY$ and let $\angle AYB=2a$ , $AXYD$ is cyclic $\implies \angle AYX=\angle ADX=a$ .Also $\angle AXB=90+a \implies \angle AXD=90-a$ .So in $\Delta AXD,\angle XAD=180-90+a+a=90=\angle CAD$

This post has been edited 1 time. Last edited by mathetillica, Apr 18, 2019, 12:42 PM

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