Question

Let ABC be a triangle and a circle C’ be drawn lying inside the triangle, touching its incircle C externally and also touching the two sides AB and AC. Show that the ratio of the radii of the two circles C’ and C is equal to \({tan}^2\big(\frac{\pi - 2}{4}\big) \).

Answer 1

Let C be the incentre, r the inradius and E the point of contact of the incircle with AB. Let C′ be the centre of the circle touching AB, AC and the incircle, r′ the radius of this circle and F its point of contact with AB. Since AB and AC both touch this circle, its centre must also lie on AC.

From C′ draw C′D ⊥ CE. Then, in ∆ C′CD 

CD = r – r′ 

CC′ = r + r′ 

∠CDC′ ⇒ π/2 and ∠DC′C = ∠EAC = A/2 

In ∆ DCC′ ⇒ sin A/2 = \(\frac{CD}{CC'}\) = \(\frac{r-r'}{r+r'}\) 

⇒ cos (π/2 – A/2) = \(\frac{r-r'}{r+r'}\)