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Saudi Arabian IMO Booklet

Geometry Difficulty 7.9 National Olympiad, round 2 Prove it Saudi Arabia

Given is an acute triangle $ABC$ with $BC < CA < AB$ . Points $K$ and $L$ lie on segments $AC$ and $AB$ and satisfy $AK = AL = BC$ . Perpendicular bisectors of segments $CK$ and $BL$ intersect line $BC$ at points $P$ and $Q$ , respectively. Segments $KP$ and $LQ$ intersect at $M$ . Prove that $CK + KM = BL + LM$ .

Solution

Let $D$ and $E$ be points of rays $ML \to$ and $MK \to$ , respectively, such that $LD = AB$ and $KE = AC$ . As $\angle DLA = \angle BLM = \angle LBC$ , $LD = AB$ and $AL = BC$ , triangles $DLA$ and $ABC$ are congruent. Analogously, $EAK$ and $ABC$ are congruent. From
$\angle DAL + \angle LAK + \angle KAE = \angle BCA + \angle CAB + \angle ABC = 180^\circ,$
it follows that $A$ belongs to the segment $DE$ . Therefore, in the light of $\angle LDA = \angle BAC = \angle AEK$ , triangle $MDE$ is isosceles with $MD = ME$ . Note that
$\begin{aligned} BL + LM &= AB + LM - AL = DL + LM - AL = DM - AL, \\ CK + KM &= AC + KM - AK = EK + KM - AK = EM - AK. \end{aligned}$
From $DM = EM$ and $AL = AK$ it follows that the right-hand sides of the two equalities above are equal. Hence the left-hand sides are equal as well, i.e. $CK + KM = BL + LM$ .

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