Question

In a triangle, the ratio of the distance between a vertex and the orthocentre and the distance of the circumcentre from the side opposite the vertex is

(a) 3 : 1 

(b) 4 : 1 

(c) 2 : 1 

(d) √2 :1

Answer 1

(c) 2 : 1

Let ABC be the given triangle whose circumcentre is O. 

Produce CO to meet the circle at G ⇒ COG is the diameter of the circle. 

⇒ ∠GAC = ∠GBC = 90º           (Angles in a semicircle) 

Let AF and BE be the perpendiculars from vertex A and B respectively on sides BC and AC. 

∴ H is the orthocentre of DABC. We need to find the ratio AH : OD, where OD is the perpendicular distance of the circumcentre O from side BC. 

OD ⊥ BC ⇒ BD = DC (Perpendicular from the centre of the circle bisects the chord) 

Also, GB ⊥ BC and OD ⊥ BC ⇒ OD || GB. 

∴ In ΔBGC, by the midpoint theorem, O and D being the mid-points of sides GC and BC, OD || GB and OD = \(\frac{1}{2}\) GB 

From (i)

 Now GA and BE are both perpendiculars to AC 

⇒ GA || BE ⇒ GA || BH 

Also, GB || AF ⇒ GB || AH 

⇒ GAHB is a parallelogram 

⇒ GB = AH = 2OD

∴ \(\frac{AH}{OD} = \frac{2OD}{OD}\) = \(\frac{2}{1}\).