Circumcircles tangent

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Circumcircles tangent

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Source: Junior Olympiad of Malaysia Shortlist 2015 G8

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#1 h Jul 17, 2015, 1:58 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCDE$ be a convex pentagon such that $BC$ and $DE$ are tangent to the circumcircle of $ACD$ . Prove that if the circumcircles of $ABC$ and $ADE$ intersect at the midpoint of $CD$ , then the circumcircles $ABE$ and $ACD$ are tangent to each other.

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#2 h Jul 17, 2015, 3:29 PM • 3 Y

Y by enhanced, Adventure10, Mango247

My solution :

Let $\tau$ be the tangent of $\odot (ACD)$ through $A$ .
Let $M$ be the midpoint of $CD$ and $T \equiv BC \cap DE$ .

From $\angle ACB=\angle ADM, \angle CBA=\angle DMA \Longrightarrow \triangle ABC \sim \triangle AMD$ .
Similarly, $\triangle AED \sim \triangle AMC \Longrightarrow \angle BAE=2\angle CAD =180^{\circ}-\angle ETB \Longrightarrow T \in \odot (ABE)$ .

Since $AT$ is the isogonal conjugate of $AM$ WRT $\angle CAD$ (well-known) ,
so $\measuredangle (\tau, AT)=\measuredangle (\tau, AC) +\measuredangle CAT=\measuredangle ADC+\measuredangle MAD=\measuredangle AMC=\measuredangle AET$ ,
hence $\tau$ is also the tangent of $\odot (ABE)$ through $A \Longrightarrow \odot (ABE)$ and $\odot (ACD)$ are tangent to each other .

Q.E.D

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#3 h Jul 17, 2015, 3:37 PM • 2 Y

Y by Adventure10, Mango247

My solution:
Let $M$ is the midpoint of $CD\Rightarrow \odot (ABC)\cap \odot (ADE)=\left \{ A,M \right \}$ .
We have $\angle DME+\angle DEM=\angle xDC=\angle CAD=\angle CAM+\angle MAD=\angle CBM+\angle DEM\Rightarrow \angle DME=\angle CBM$ .
Similarly $\angle DEM=\angle CMB$ . So $\triangle MBC\sim \triangle EMB$ (a.a) $\Rightarrow \frac{MB}{ME}=\frac{MC}{DE}=\frac{MD}{DE}$ , and $\angle BME=180^{\circ}-(\angle BMC+\angle DME)=180^{\circ}-(\angle MED+\angle DME)=\angle MDE$ $\Rightarrow \triangle BME\sim \triangle MDE$ (s.a.s) $\Rightarrow \angle BEM=\angle MED=\angle BMC$ $(*)$
Let $At$ be the tangent line of $\odot (ACD)$ , we have $\angle tAC=\angle ADC\Leftrightarrow \angle tAB+\angle BAC=\angle AEM=\angle AEB+\angle BEM=\angle AEB+\angle BMC$ ( $\because (*)$ ) $=\angle AEB+\angle BAC$ $\Leftrightarrow \angle tAB=\angle AEB$ $\Rightarrow At$ is tangent to $\odot (ABE)$ $\Rightarrow \odot (ACD)$ is tangent to $\odot (ABE)$ . DONE
Remark: Let $BC\cap DE=F$ then we prove that $M$ is the incenter of $\triangle BEF$ , and combine with $\odot (ACD)$ is tangent to $FB, FE$ $\Rightarrow \odot (ACD)$ is the F-mixtilinear incircles of $\triangle BEF$ $\Rightarrow \odot (ACD)$ is tangent to $\odot (ABE)$ .

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This post has been edited 1 time. Last edited by LeVietAn, Jul 17, 2015, 3:39 PM

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#4 h Aug 27, 2023, 1:57 PM

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Angle chasing... Let $BC\cap DE=R$ and midpoint of $CD$ is $M$ . Then $AR$ is symmedian of $\angle ACD$ . Then we have $\angle CAB=\angle CMB=\angle RCD-\angle CBM=\angle CAD-\angle CAM=\angle MAD$ , similarly $\angle DAE=\angle CAM$ , which means that $\angle BAE=2\angle CAD$ adding to this $\angle BRE= 180-2\angle CAD$ gives that $(R B A E)$ is cyclic. Since $\angle DAM=\angle RAC$ (because $AR$ is symmedian) we have $\angle MEB=\angle DEB-\angle DEM=\angle RAB-\angle DAM= \angle DAM$ . Then let $AA$ be the tangent to $(A C D)$ and $K$ be the point on $AA$ , then $KAB=\angle KAC-\angle BAC=\angle MDA-\angle MAD=\angle MEA-\angle MEB=\angle BEA$ , which completes the solution.

This post has been edited 1 time. Last edited by X.Allaberdiyev, Aug 27, 2023, 1:57 PM
Reason: Latex

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