Question

Let ABC be an acute-angled triangle, let D, F be the midpoints of BC, AB respectively, let the perpendicular from F to AC and the perpendicular at B to BC meet in N. Prove that ND is equal to the circum-radius of ABC.

Answer 1

Explanation:

Let O be the circumcontre of ABC. Join OD, ON and OF. We show that BDON is a rectangle. It follows that DN = BO = R, the circum-radius of ABC.

Observe that ∠NBC = ∠NKC = 90°. Hence BCKN is a cyclic quadrilateral.Thus ∠KNB = 180° - ∠BCA. But ∠BOA = 2∠BCA and OF bisects ∠BOA. Hence ∠BOF = ∠BCA. We thus obtain

∠FNB + ∠BOF = ∠KNB + ∠BCK = 180°

This implies that B, O, F, N are con-cyclic. Hence ∠BFO = ∠BNO. But observe that ∠BFO = 90°. Since OF is perpendicular to AB. Thus ∠BNO = 90°. Since NB and OD are perpendicular to BC, It follows that BDON is rectangle.

Alternate solution: