# Let $T_1$ and $T_2$ be two circumferences…

Let $T_{1}$ and $T_{2}$ be two circumferences with centers $O_{1}$ and $O_{2}$ respectively, such that $T_{1}$ passes through $O_{2}.$ Let $C$ be a point on $T_1$. Let $r_{1}$ and $r_{2}$ be the lines tangent to $T_{2}$ that pass through $C,$ and let $A$ and $B$ be the points where $r_{1}$ and $r_{2}$ cut $T_{1}$ again.

Here is a drawing.

Problem: Show that $AB$ is perpendicular to $O_{1}O_{2}.$

The radical axis of $T_{1}$ and $T_{2}$ is perpendicular to $O_{1}O_{2},$ so we want to show that $AB$ is parallel to the axis, but I have not been able to continue from there.

By symmetry, the two tangents from $C$ to $T_2$ are the same angle from $CO_2$, i.e. angle $BCO_2$ equals angle $ACO_2$. On circle $T_1$, these are the inscribed angles of arcs $AO_2$ and $BO_2$, which must therefore be the same. Since angle $AO_1O_2$ equals angle $BO_1O_2$, triangles $AO_1D$ and $BO_1D$ are congruent, which implies $O_1D \perp AB$.

From the green circle, y = z.

From the grey circle, x = z.

Then, x = y. Done.

Added: AB, MN, and the line of centers will be concurrent at T.

Suppose that AB cuts MN at T; and $O_2T$ is joined.

Let $O_1T$ and $TO_2$ be two straight lines joined at T only.

Then, the result of $x = y = z$ is still valid. This means $\angle ATO_2 = \angle BTO_2 = 90^0$.

In turn, $BNTO_2$ is cyclic with $u = v$. But $v = x$ (radii of the same (red) circle).

Then, $\triangle O_2TA \cong \triangle O_2TB$. this further means $O_2T$ is the perpendicular bisector of $AB$.

Note that $O_1T$ is the line joining the center to the midpoint (T) of its chord (AB). This makes $O_2TO_1$ is a straight line.

• How did you prove that M, N, and the intersection of AB and $O_1O_2$ are collinear? – problembuster Jul 28 '16 at 8:13
• @problembuster See added. – Mick Jul 28 '16 at 16:16
• @problembuster it is called the simson-wallace line. – Weijie Chen Jul 28 '16 at 16:26

It's very simple just notice this very useful lemma in math contests:

Lemma 1: The angle bisector of the vertex $A$ of a triangle $ABC$ and the perpendicular bisector meets at the midpoint of the arc $BC$ of the circumcircle $ABC$ without containing $A$. The proof if angle chasing I let it for you.

Now as the problem states, $CA$ and $CB$ are tangent to $T_2$ hence $CO_2$ is the angle bisector of $CB$ and $CA$. Now clearly by lemma 1 $O_1O_2$ will be perpendicular to $BA$ moreover it passes through the midpoint of the segment $AB$.