MathDB

In the triangle

ABC

, the inscribed circle touches the sides

CA{}

and

AB{}

at the points

B_1{}

and

C_1{}

, respectively. An arbitrary point

D{}

is selected on the side

AB{}

. The point

L{}

is the center of the inscribed circle of the triangle

BCD

. The bisector of the angle

ACD

intersects the line

B_1C_1

at the point

M{}

. Prove that

\angle CML=90^\circ

.
Proposed by Chan Quang Heung (Vietnam)

M2615