Similar triangles ACD, AEF, AMN

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Similar triangles ACD, AEF, AMN

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Source: German Mathematical Competition BWM 2005, 2nd round, problem

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#1 h Sep 1, 2005, 10:03 PM • 2 Y

Y by Adventure10, Mango247

Two circles $k_1$ and $k_2$ intersect at two points $A$ and $B$ . Some line through the point $B$ meets the circle $k_1$ at a point $C$ (apart from $B$ ), and the circle $k_2$ at a point $E$ (apart from $B$ ). Another line through the point $B$ meets the circle $k_1$ at a point $D$ (apart from $B$ ), and the circle $k_2$ at a point $F$ (apart from $B$ ). Assume that the point $B$ lies between the points $C$ and $E$ and between the points $D$ and $F$ .

Finally, let $M$ and $N$ be the midpoints of the segments $CE$ and $DF$ .

Prove that the triangles $ACD$ , $AEF$ and $AMN$ are similar to each other.

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ThAzN1

#2 h Sep 1, 2005, 10:25 PM • 2 Y

Y by Adventure10, Mango247

First, $\triangle ACD \sim \triangle AEF$ follows from

$\angle CAD = \angle CBD = \angle FBE = \angle FAE,$
$\angle ACD = \pi - \angle ABD = \angle ABF = \angle AEF.$

Now, note that $\triangle ACE \sim \triangle ADF$ because

$\angle ACE = \angle ACB = \angle ADB = \angle ADF,$
$\angle AEC = \angle AEB = \angle AFB = \angle AFB.$

Thus, $\triangle ACM \sim \triangle ADN$ and we have

$\frac{AM}{AN} = \frac{AC}{AD}.$

Additionally, $\triangle ACE$ is taken to $\triangle ADF$ by a spiral similarity about $A$ , with rotation angle $\angle CAD$ , ratio $AD/AC$ so $M$ goes to $N$ under this spiral similarity and $\angle MAN = \angle CAD$ , so the result follows...

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jayme

#3 h Sep 24, 2013, 10:33 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
according to the midcircle theorem, A, B, M and N are concyclic.
By an angle chasing, we are done.
Sincerely
Jean-Louis

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