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Geometry Difficulty 4.9 AIME Prove it China

As shown in the figure, BCBC and ACAC are tangent to the inscribed circle II of △ABC\triangle ABC at MM and NN. EE and FF are the midpoints of ABAB, ACAC respectively. EFEF intersects BIBI at DD. Prove that MM, NN, DD are collinear.

Figure 1

Solution

Join ADAD, and it is obvious that ∠ADB=90∘\angle ADB = 90^\circ. Then join AIAI and DMDM. Suppose DMDM intersects ACAC at GG. Since ∠ABI=∠DBM\angle ABI = \angle DBM, we obtain ABBD=BIBM\frac{AB}{BD} = \frac{BI}{BM}. Hence, △ABI∼△DBM\triangle ABI \sim \triangle DBM, and
∠DMB=∠AIB=90∘+12∠ACB. \angle DMB = \angle AIB = 90^\circ + \frac{1}{2} \angle ACB.
Join IGIG, ICIC, IMIM, then
∠IMG=∠DMB−90∘=12∠ACB=∠GCI. \angle IMG = \angle DMB - 90^\circ = \frac{1}{2} \angle ACB = \angle GCI.
Therefore, II, MM, CC, GG are concyclic, and IG⊥ACIG \perp AC.
Consequently, since GG and NN represent the same point, we conclude that MM, NN, DD are collinear.

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