Problem 5

Let

I

be the incenter of triangle

ABC

. Let the incircle of

ABC

touch the sides

BC

CA

, and

AB

K

L

, and

M

, respectively. The line through

B

parallel to

MK

meets the lines

LM

and

LK

R

and

S

, respectively. Prove that angle

RIS

is acute.

Step 3 of 4: Compute the product on the line

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In plain words

The half-angle factors cancel.

BR\cdot BS=BM\cdot BK=BK^2

Detailed analysis

Angle chasing in triangles $BRS$ , $BML$ , and $BKL$ (or the sine rule) gives $BR/BM=\cos(A/2)/\cos(C/2)$ and $BS/BK=\cos(C/2)/\cos(A/2)$ . Hence $BR\cdot BS=BM\cdot BK$ . Equal tangent lengths from $B$ give $BM=BK$ , so the product is $BK^2$ .