MathLabs

Problem 5

Let I I be the incenter of triangle ABC ABC . Let the incircle of ABC ABC touch the sides BC BC , CA CA , and AB AB at K K , L L , and M M , respectively. The line through B B parallel to MK MK meets the lines LM LM and LK LK at R R and S S , respectively. Prove that angle RIS RIS is acute.
Step 2 of 4: Use the cosine-law criterion
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In plain words

An angle is acute when the opposite cosine-law expression is positive.

RI2+SI2−RS2=2BI2−2BR⋅BSRI^2+SI^2-RS^2=2BI^2-2BR\cdot BS
Detailed analysis

The right triangles with altitude BI BI give RI2=BR2+BI2 RI^2=BR^2+BI^2 and SI2=BS2+BI2 SI^2=BS^2+BI^2. Since R,B,S R,B,S are collinear, RS=BR+BS RS=BR+BS . Subtracting yields RI2+SI2−RS2=2BI2−2BR⋅BS RI^2+SI^2-RS^2=2BI^2-2BR\cdot BS .