Prove that $GH=IJ$.

Question

I am given this graph:

And the problem is given as: Let us take commont exterior tangent lines $\ell_1$ and $\ell_2$ for two disjoint cirles on the plane. Suppose that $\ell_1$ touches $\omega_1$ at the point $G$, $\ell_2$ touches $\omega_2$ at the point $I$. Suppose that the segment $GI$ intersects $\omega_1$ and $\omega_2$ at the points $H$ and $J$, respectively. Prove that $GH$ = $IJ$.

I have no idea how the conclusion is derived. Any help please?

Answer 1

Consider the points $A$ and $B$ marked on the figure.

Using the concept of "Power of Point" into the circumference on the left side, we have:

$$ IH \cdot IG = IA^2$$ (1)

Similarly, on the circumference on the right side

$$ GJ \cdot IG = GB^2$$ (2)

Note that $GB = AI$, then from (1) and (2), we have:

$$ GJ \cdot IG = IH \cdot IG $$

$$ GJ = IH $$

$$ GH + HJ = IJ + HJ $$

$$ GH = IJ$$

qed.