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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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Answer» 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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