CGMO, 2022

Geometry Difficulty 4.3 AIME Prove it China

As shown in the figure, points $K$ and $L$ are in the interior of a triangle $ABC$ , point $D$ lies on the side $AB$ . It is known that points $B$ , $K$ , $L$ , $C$ are concyclic, and $\angle AKD = \angle BCK$ , $\angle ALD = \angle BCL$ . Prove that $AK = AL$ .

Solution

Proof. As shown in the following picture,

let the extensions of $AK$ and $AL$ intersect the circle passing through $B$ , $K$ , $L$ , $C$ at points $X$ and $Y$ , respectively. Connect $BX$ and $BY$ . Considering the given conditions, we have $\angle AKD = \angle BCK = \angle BXK$ and $\angle ALD = \angle BCL = \angle BYL$ .

Thus, $DK \parallel BX$ and $DL \parallel BY$ . So we have
$\frac{AK}{AX} = \frac{AD}{AB} = \frac{AL}{AY}.$
Since $K$ , $L$ , $Y$ , $X$ are concyclic, by the Power of a Point theorem, we have
$AK \cdot AX = AL \cdot AY.$
Multiplying the two equations above, we get $AK^2 = AL^2$ , which implies $AK = AL$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.