Maths

圖片參考：https://s.yimg.com/rk/HA04628698/o/893682127.jpg


Let N be a point such that BM = MN = NB , then ∠BMN = 60° (Property of equil. Δ)
By given, circles O and O' with same radius, then ΔAOB ≌ ΔAO'B (S.S.S.)
Hence ∠AMB
= ½∠AO'B (∠at centre twice ∠ at ☉ce)
= ½∠AOB = 30° (Corresponding angles of congruent triangle)
We have ∠AMN =∠BMN +∠AMB = 60° + 30° = 90°
∴ AM² + MN² = AN² (Converse of Pythagoras' theorem)

Obviously, ΔAOB is an equilateral Δ, so AB = OB ;
∠ABO = 60°=∠MBN ( property of equil. Δ)
∴∠ABN = ∠ABM +∠ABO = ∠ABM +∠MBN = ∠OBM ;
NB = MB (Assumption)
Hence ΔNBA ≌ ΔMBO (S.A.S.) , we have AN = OM
and by assumption that MN = BM.
∴ AM² + MN² = AN²
⇒ AM² + BM² = OM²

2014-10-21 11:12:40 補充：
∴∠OBM = ∠ABM +∠ABO = ∠ABM +∠MBN = ∠ABN ;