<p>Let ABC be a triangle and let BB1, CC1 be respectively the bisectors of ∠B, ∠C with B1 on AC and C1 on AB. Let E, F be the feet of perpendiculars drawn from A onto BB1, CC1 respectively. Suppose D is the point at which the incircle of ABC touches AB. Prove that AD = EF. (RMO 2011)<br /> Please give solution with figure.</p>
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