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Circles $C_1$ and $C_2$ touch externally at a point $M$ and also touch a circle $C$ internally at $L_1$and $L_2$ respectively. Let $X$ be a point of intersection of $C$ with the common tangent to $C_1$ and $C_2$ at $M$. Line X $L_1$ meets $C_1$ again at $A_1$ and X $L_2$ meets $C_2$ again at $A_2$.

Prove $A_1A_2$ is a common tangent to $C_1$ and $C_2$.

All help welcome. Really struggling with this one.

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closed as off-topic by projectilemotion, Shailesh, Leucippus, mrp, Simply Beautiful Art Feb 4 '17 at 18:18

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  • $\begingroup$ Upload a picture. $\endgroup$ – Mick Feb 3 '17 at 13:37
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We must show that line $A_1A_2$ is perpendicular to both $C_1A_1$ and $C_2A_2$.

By the tangent secant theorem we have $XA_1\cdot XL_1=XM^2=XA_2\cdot XL_2$. It follows that triangles $XA_1A_2$ and $XL_2L_1$ are similar, so that (see diagram below for angle names): $$ \angle XA_1A_2=\angle XL_2L_1={\pi\over2}-\alpha+\gamma, \quad \angle XA_2A_1=\angle XL_1L_2={\pi\over2}-\alpha+\beta. $$ In addition, notice that: $$\beta+\gamma=\angle L_1XL_2={1\over2}\angle L_1CL_2=\alpha.$$ We have then: $$ \angle C_1A_1A_2=\pi-\angle C_1A_1L_1-\angle XA_1A_2= \pi-\beta-\left({\pi\over2}-\alpha+\gamma\right)={\pi\over2}, $$ as it was to be proved. A similar reasoning can be made for $\angle C_2A_2A_1$.

enter image description here

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Let's look at the figure below.

enter image description here

Construct the common tangent line (blue). Denote the touching points by $A_1$ and $A_2$. Then drop perpendiculars from $C_1$ and from $C_2$ to the blue line. Then take the line connecting $L_1$ and $A_1$ and the line connecting $L_2$ and $A_2$. These two lines will meet on $C$ at $X$.

Certain triangles and the small circles on the figure are homothetic images of each other. The homothetic center is $H$. As a result the triangles $L_1A_1D_1$ and $MA_2D_2$ and the triangles $L_1A_1C_1$ and $MA_2C_2$ are similar triangles.

Comparing the angles of these triangles with the corresponding angles of the big triangle $L_1XL_2$ will show that, say, $L_1A_1D_1$ and $L_1XL_2$ are similar. It turns out that the angle at $X$ is a right angle. (Why are the angles at $D_2$ and $L_2$ equal?) That is, the big black triangle is a Thales triangle of the red circle. This is why is $X$ on the red circle.

This proves that doing the construction in the revers order we will find $A_1$ and $A_2$ on the tangent line.

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