MathLabs

Problem 1

Let MM be a point on side ABAB of triangle ABCABC. Let r1,r2,rr_1,r_2,r be the inradii of triangles AMCAMC, BMCBMC, and ABCABC, respectively. Let q1,q2,qq_1,q_2,q be the radii of the excircles of these triangles lying in angle ∠ACB\angle ACB. Prove that r1q1⋅r2q2=rq\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac rq.
Step 5 of 5: Use supplementary angles at MM
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In plain words

The two angles at the cut point fill a straight angle, so their half-angle factors cancel.

tan⁡∠AMC2tan⁡∠BMC2=1\tan\frac{\angle AMC}{2}\tan\frac{\angle BMC}{2}=1
Detailed analysis

Because A,M,BA,M,B are collinear, ∠AMC+∠BMC=π\angle AMC+\angle BMC=\pi. Hence the two half-angles are complementary, so their tangents are reciprocal. Multiplying the two formulas from the preceding step gives r1q1r2q2=tan⁡A2tan⁡B2=rq\frac{r_1}{q_1}\frac{r_2}{q_2}=\tan\frac A2\tan\frac B2=\frac rq.