# Circle and Tangents Geometry proof Problem [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – projectilemotion, Shailesh, Leucippus, mrp, Simply Beautiful Art
If this question can be reworded to fit the rules in the help center, please edit the question.

• Upload a picture. – Mick Feb 3 '17 at 13:37

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 $\angle C_2A_2A_1$.

Let's look at the figure below.

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.