Cyclic Quadrilaterals Geometry

Geometry Level 3

A circle with center $O$ passes through the vertices $A$ and $C$ of triangle $ABC$ and intersects segments $AB$ and $BC$ again at distinct points $K$ and $N$ , respectively. The circumcircles of triangles $ABC$ and $KBN$ intersects at exactly two distinct points $B$ and $M$ . What is the value of $\angle OMB$ ?

90 45 60 75

1 solution

Rohan Chandra
May 14, 2015

The lines AC, KN, BM concur at the radical center P of the three circles involved. Let A, B, P, Q be four distinct points on a plane. Then AB ⊥ PQ if and only if $PA^{2}$ - $PB^{2}$ = $QA^{2}$ - $QB^{2}$

the quadrilateral PCNM is cyclic since ∠PCN = ∠ AKN = ∠BMN. By intersecting chords theorem we have , PM x PB = PC x PA = $OP^{2}$ - $r^{2}$ [r is the circumradius of triangle AKC] BM x BP = BN x BC = $OB^{2}$ - $r^{2}$

Therefore, ) 1. $OB^{2}$ - $OP^{2}$ = BM x BP - PM x PB 2. BP x (BM - PM) 3. (BM + PM) x (BM + PM) 4. $BM^{2}$ - $PM^{2}$

Hence, OM ⊥ BP ; ∠OMB = $90^{2}$

Cyclic Quadrilaterals Geometry

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