In the Solution above, I claim that HG is parallel to IE. The way to see that is to use a homothety, dilating about the point D, that expands the small circle into the large circle. Under this homothety, the points I and E on the small circle move to the points H and G, respectively, on the large circle, the secant IE expands to the secant HG, and the secants are parallel.

Ignacio Larrosa CaÃ±estro pointed out that

“If the circles aren’t tangent and the common internal tangent is replaced by their radical axis, the result remains.”

The Solution above also can be used to prove this generalization.

He also pointed out (implicitly) that it is not necessary to invoke the 3 radical axes theorem. Instead, the cyclic quadrilateral condition can be used to show that the intersection point J of the lines HF and IE must have the same power with respect to both the big and small circles, implying the J must be on the radical axis.

## 2 replies on “On a tangent”

In the Solution above, I claim that HG is parallel to IE. The way to see that is to use a homothety, dilating about the point D, that expands the small circle into the large circle. Under this homothety, the points I and E on the small circle move to the points H and G, respectively, on the large circle, the secant IE expands to the secant HG, and the secants are parallel.

Ignacio Larrosa CaÃ±estro pointed out that

“If the circles aren’t tangent and the common internal tangent is replaced by their radical axis, the result remains.”

The Solution above also can be used to prove this generalization.

He also pointed out (implicitly) that it is not necessary to invoke the 3 radical axes theorem. Instead, the cyclic quadrilateral condition can be used to show that the intersection point J of the lines HF and IE must have the same power with respect to both the big and small circles, implying the J must be on the radical axis.