cyclic quadrilateral starting with 2 intersecting circles (midpoint, tangents)

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cyclic quadrilateral starting with 2 intersecting circles (midpoint, tangents)

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Source: Australian MO 2017

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#1 h Sep 23, 2018, 5:56 PM • 2 Y

Y by Adventure10, Mango247

The circles $K_1$ and $K_2$ intersect at two distinct points $A$ and $M$ . Let the tangent to $K_1$ at $A$ meet $K_2$ again at $B$ , and let the tangent to $K_2$ at $A$ meet $K_1$ again at $D$ . Let $C$ be the point such that $M$ is the midpoint of $AC$ . Prove that the quadrilateral $ABCD$ is cyclic.

This post has been edited 2 times. Last edited by parmenides51, Sep 23, 2018, 8:35 PM
Reason: latex

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#2 h Sep 23, 2018, 10:39 PM • 1 Y

Y by Adventure10

Take inversion with center $A$ and radius $r$ .
Now $AB'M'D'$ is a parallelogram and $C'$ is the midpoint of $AM'$ . So $D'$ , $B'$ and $C'$ are collinear $\Longrightarrow ABCD$ is cyclic. So we are done.
$\blacksquare$

This post has been edited 1 time. Last edited by Anaskudsi, Sep 23, 2018, 10:40 PM

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#3 h Mar 30, 2020, 1:09 PM • 1 Y

Y by gamerrk1004

Let $C' = AM \cap \odot (ABD)$ . Since, $AD$ and $AB$ are tangents to the respective circles, $\implies$ $\tfrac{AD}{AB}$ $=$ $\tfrac{AM}{BM}$ $=$ $\tfrac{DC'}{C'B}$ $\implies$ $AC'$ is the $A-$ symmedian of $\Delta ABD$ and since, $DM$ , $DB$ are Isogonal WRT $\angle ADC'$ $\implies$ $C=C'$ $\qquad \blacksquare$

This post has been edited 1 time. Last edited by AlastorMoody, Mar 30, 2020, 1:15 PM

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