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Turkey — Team Selection Test

Geometry Difficulty 8.4 Shortlist Prove it Turkey

Let a line $l$ intersect the line $AB$ at $F$ , the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$ , respectively and the internal bisector of the angle $BAC$ at $P$ . Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$ , $CD = CE$ and $P$ is in the interior the triangle $ABC$ . Prove that
$FB \cdot FA + CP^2 = CF^2 \iff AD \cdot BE = PD^2.$

Solution

Let $DP = a$ , $PE = b$ , $EF = c$ , $CD = CE = x$ . By the Stewart Theorem $CP^2 = x^2 - ab$ and $CF^2(a+b) + x^2c = (a+b+c)((a+b)c + x^2)$ , hence $CF^2 - CP^2 = (a+c)(b+c)$ . Let $Q$ be a point on the line segment $DE$ satisfying $EQ = a$ . Then $CF^2 - CP^2 = (a+c)(b+c) = FP \cdot FQ$ . If $CF^2 - CP^2 = FB \cdot FA$ , then we get $FB \cdot FA = FP \cdot FQ$ . Hence the points $A, B, P, Q$ are concyclic and $\angle BQE = \angle BAP = \angle DAP$ . Since $\angle QEB = \angle ADP$ , we get that the triangles $ADP$ and $QEB$ are similar and hence $AD \cdot BE = PD \cdot QE = PD^2$ . If $AD \cdot BE = PD^2 = PD \cdot QE$ , then the triangles $ADP$ and $QEB$ are similar and hence $\angle EQB = \angle DAP = \angle PAB$ and therefore the points $A, B, P, Q$ are concyclic. Hence $FB \cdot FA = FP \cdot FQ = (a+c)(b+c) = CF^2 - CP^2$ ,

and we are done.

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