Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it Ukraine

In a triangle $ABC$ on the side $AC$ the points $F$ and $L$ with $AF = LC < \frac{1}{2}AC$ are chosen. Find an angle $FBL$ , if $AB^2 + BC^2 = AL^2 + LC^2$ .

Answer: $\angle FBL = 90^\circ$ .

Solution

Let $AB = c$ , $BC = a$ , $FM = ML = c$ ( $M$ is a middle of $AC$ ), $AF = LC = x$ . According to known formula find length of median (Fig.25)

Fig.25

$\begin{aligned} BM^2 = m^2 &= \frac{1}{4}(2c^2 + 2a^2 - (2b + 2x)^2) = \frac{1}{4}(2(2b + x)^2 + 2x^2 - (2b + 2x)^2) = \\ &= \frac{1}{4}(8b^2 + 8bx + 4x^2 - 4b^2 - 8bx - 4x^2) = b^2. \end{aligned}$

Therefore in the $\triangle FBL$ median is half of base, in which it has marked, that's why this triangle is right and $\angle FBL = 90^\circ$ .

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