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 4 of 4: Conclude acuteness
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In plain words

The strict positivity comes from the inradius segment IK IK .

RI2+SI2−RS2=2IK2>0RI^2+SI^2-RS^2=2IK^2>0
Detailed analysis

Substituting the product gives RI2+SI2−RS2=2(BI2−BK2)=2IK2>0 RI^2+SI^2-RS^2=2(BI^2-BK^2)=2IK^2>0, since BIK BIK is right at K K . By the converse of the cosine rule, ∠RIS\angle RIS is acute.