Problem 5

A circle with center

O

passes through vertices

A

and

C

of triangle

ABC

and intersects segments

AB

and

BC

again at distinct points

K

and

N

, respectively. The circumcircles of

ABC

and

KBN

meet at exactly two distinct points

B

and

M

. Prove that

\angle OMB

is a right angle.

Step 4 of 4: Subtract and recognize perpendicularity

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In plain words

The final distance identity is a disguised Pythagorean statement: after expanding squared distances from $M$ , the cross term is zero, which means the two directions are perpendicular.

XM^2-BM^2=XO^2-BO^2

Detailed analysis

Subtract the identities in Step 3: $XM\cdot XB-BM\cdot BX=XO^2-BO^2$ . Since $X,M,B$ are collinear, the left side is $XB(XM-BM)=XM^2-BM^2$ . Thus $XO^2-XM^2=BO^2-BM^2$ . In vector form with origin at $M$ , this says $|\vec{MO}+\vec{MX}|^2-|\vec{MX}|^2=|\vec{MB}+\vec{MX}|^2-|\vec{MX}|^2$ , hence $\vec{MO}\cdot\vec{MB}=0$ ; therefore $OM\perp BM$ and $\angle OMB=90^\circ$ .