|CA| |CD| = |AB| |AM| , angle bisectors, reflection of circumcenter related

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|CA| |CD| = |AB| |AM| , angle bisectors, reflection of circumcenter related

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Source: Dutch NMO 2019 p3

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#1 h Jan 9, 2020, 11:34 AM • 1 Y

Y by Adventure10

Points $A, B$ , and $C$ lie on a circle with centre $M$ . The reflection of point $M$ in the line $AB$ lies inside triangle $ABC$ and is the intersection of the angle bisectors of angles $A$ and $B$ . Line $AM$ intersects the circle again in point $D$ .
Show that $|CA| \cdot |CD| = |AB| \cdot |AM|$ .

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tk1

#2 h Jan 9, 2020, 3:54 PM • 1 Y

Y by Adventure10

Let $I$ be the incenter. By construction, $\angle AMB=\angle AIB,$ and thus $360^\circ-2\angle C=90^\circ +\frac{\angle C}{2},$ or $\angle C=108^\circ.$ Therefore $\angle A+\angle B=72^\circ;$ but by construction, $\angle IAB=\angle BAM=\angle MBA=\angle ABI,$ and therefore $\angle A=\angle B=36^\circ,$ $\angle DAC=54^\circ,$ and $\angle AMB=144^\circ.$ Now $CA\cdot CD=2R\sin 36^\circ\cdot 2R\sin 54^\circ =4R^2\sin 36^\circ\cos 36^\circ=2R^2\sin 72^\circ=2R\sin 72^\circ\cdot R=AB\cdot AM.$

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#3 h Dec 31, 2020, 2:01 PM

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Kind of easier. Let $I$ be the incenter of $\triangle ABC$ . Since $CI$ and $IM$ both go through the midpoint of arc $AB$ (since $CI$ is the angle bisector as well and $MA=MB$ ) we get that $C,I$ and $M$ are collinear. This means that $CA=CB$ . So now since $\angle MDC= \angle B$ , $MC=MD$ and $CA=CB$ we get that $\triangle CAB\sim \triangle MCD$ and so,
$\tfrac {CD}{AM}=\tfrac {CD}{DM}=\tfrac {AB}{CA}$
and we are done.

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Leo890

#4 h Jan 8, 2021, 11:36 PM

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Because $I$ is the reflection of $M$ about $AB$ and $M$ is on the perpendicular bisector of $AB$ , thus $AI=IB$ . Define $\angle CAI = \alpha$ . Thus, $\angle CAI = \alpha = \angle IAB = \angle IBA = \angle IBC$ . It follows that $CA = CB$ . Now $\frac {CA}{AB} = \frac{CB}{AB} = \frac {sin(2\alpha)} {sin(4\alpha)}$ by Law of sines on $\triangle ABC$
Similarly $\angle CMD = 2 \angle CAD = 180^\circ - 4\alpha$ . Thus, Law of sines on $\triangle CMD$ gives $\frac{AM}{CD} = \frac{MD}{CD} = \frac {sin(2\alpha)}{sin(4\alpha)} =\frac{CB}{AB}$ , and the conclusion follows. =)

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