SELECTION and TRAINING SESSION

Question

Let be a circle, S be a point outside and SA, SB be tangents from S to . Arbitrary point K is chosen on the segment AB. Points U, W, R and T are chosen on such that UW RT = K and the points W, T lie on the same half-plane with respect to the line AB. The line SK intersects at P and Q and P lies on the same half-plane to the line AB as W and T. The lines WR and TU intersect AB at C and D respectively. Let M be the midpoint of PQ. Prove that AMC = BMD.

Accepted Answer

We will use multiple times the following well-known Lemma. Let the chords

A B

and

C D

of a circle intersect at the point

K

then

\frac{A C \cdot A D}{B C \cdot B D} = \frac{C K}{D K}

Proof of lemma. This equality directly follows from the sine laws for the triangles

A C K, A D K

and

B C D

.

The quadrilateral

A P B Q

is harmonic, so

M K

is the bisector of the angle

A M B

. Hence

\frac{A C}{C B} = \frac{s i n A M C}{s i n C M B} \cdot \frac{A K}{B K}

and

\frac{A D}{D B} = \frac{s i n A M D}{s i n D M B} \cdot \frac{A K}{B K}

, therefore it's enough to prove the equality

\frac{A C}{C B} \cdot \frac{A D}{D B} = (\frac{A K}{B K})^{2}

.

The lemma for

A B \cap R W = C

and

A B \cap U T = D

implies that

\frac{A C}{C B} = \frac{A W \cdot A R}{B W \cdot B R} and \frac{A D}{D B} = \frac{A T \cdot A U}{B T \cdot B U}

And the lemma for

A B \cap R T = K

and

A B \cap U W = D

implies that

\frac{A K}{K B} = \frac{A T \cdot A R}{B T \cdot B R} = \frac{A W \cdot A U}{B W \cdot B U}

Combining these equalities we obtain the required equality.