Let ABC be a right triangle with ∠B = 90º. Let E and F be respectively the mid-points of AB and AC.
Let ABC be a right triangle with ∠B = 900. Let E and F be respectively the mid-points of AB and AC. Suppose the incentre I of triangle ABC lies on the circumcircle of triangle AEF. Find the ratio BC/AB.
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Draw ID ⊥ AC. Then ID = r, the inradius of ΔABC. Observe EF || BC and hence ∠AEF = ∠ABC = 900. Hence ∠AIF = 90º. Therefore ID2 = FD x DA. If a > c, then FA > DA and we have
DA = s a; and FD = FA DA = b/2 - (s - a).
Thus we obtain
r2 = (b + c - a)(a - c)/4.
But r = (c + a - b)/2. Thus we obtain
(c + a - b)2 = (b + c - a)(a - c).
Simplication gives 3b = 3a + c. Squaring both sides and using b2 = c2 + a2, we obtian 4c = 3a. Hence BC=BA = a=c = 4/3.
(If a ≤ c, then I lies outside the circumcircle of AEF).
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