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Berkeley Math Circle Monthly Contest 7

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABC$ be a triangle with incenter $I$ . A line through $I$ parallel to $BC$ intersects sides $AB$ and $AC$ at $D$ and $E$ respectively. Prove that the perimeter of $\triangle ADE$ is equal to $AB + AC$ .

Solution

Solution:

Because $DE \parallel BC$ , $\angle BID = \angle IBC$ which is the same as $\angle DBI$ since $I$ is on the bisector of $\angle ABC$ . Thus $\triangle BDI$ is isosceles, implying $BD = DI$ . Similarly $CE = EI$ . Thus the perimeter of $\triangle ADE$ is
$AD + AE + DE = AD + AE + DI + EI = AD + AE + BD + CE = AB + AC$

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