Problem 2

Let

ABC

be a triangle and let points

M

and

N

be the midpoints of sides

AB

and

AC

, respectively. Let points

K

and

L

be chosen strictly inside triangles

BMC

and

BNC

, respectively, such that

K

lies strictly inside triangle

ABL

and

L

lies strictly inside triangle

AKC

. Suppose that

\angle KBA=\angle ACL

\angle LBK=\angle LNC

, and

\angle LCK=\angle BMK

. Let

O

be the circumcentre of triangle

AKL

. Prove that

OM=ON

Step 1 of 4: Translate the three angle conditions into cyclic quadrilaterals

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X:=BK\cap AC,\quad Y:=CL\cap AB\ \Longrightarrow\ BYXC,\ MYXN,\ XNLB\ (\omega_B),\ YMKC\ (\omega_C)\ \text{cyclic}

Detailed analysis

Work with directed angles and directed lengths, so extensions require no separate cases. Define $X:=BK\cap AC$ and $Y:=CL\cap AB$ . The first hypothesis gives $\angle YBC=\angle KBA=\angle ACL=\angle XCB$ , so $BYXC$ is cyclic by the equal-angle criterion. The midpoint relation $MN\parallel BC$ (midline of $\triangle ABC$ ) and the secant-product identity for the first circle imply that the two products on the lines through M,Y and N,X are equal, hence $MYXN$ is cyclic. Next, $\angle LBX=\angle LBK=\angle LNC=\angle LNX$ shows $XNLB$ is cyclic; call its circumcircle $\omega_B$ . Symmetrically, $\angle YCK=\angle LCK=\angle BMK=\angle YMK$ shows $YMKC$ is cyclic; call its circumcircle $\omega_C$ . The strict interior hypotheses keep the needed points distinct, and directed angles cover either possible order on each line.