Problem 5

ABC

is a right-angled triangle with right angle at

A

, and

AD

is the altitude to the hypotenuse. The line joining the incenters of

ABD

and

ACD

meets

AB,AC

K,L

. If

S

and

T

are the areas of

ABC

and

AKL

, prove that

S\ge2T

Step 2 of 4: Both incenters lie on the comparison line

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\angle AK'I_1=45^o=\angle ADI_1,\qquad \angle AL'I_2=45^o=\angle ADI_2

Detailed analysis

Let I_1 and I_2 be the incenters of ABD and ACD. From the congruence above, I_1 lies on K'L' and the angle bisector at K' gives $\angle AK'I_1=45^o=\angle ADI_1$ . The analogous construction on ACD gives $\angle AL'I_2=45^o=\angle ADI_2$ , so I_2 also lies on K'L'. Therefore the line joining I_1 and I_2 is K'L', and its intersections satisfy K=K', L=L'.