Perpendicularity

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Perpendicularity

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Source: Indian RMO 1995 Problem 1

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1592 posts

Rushil

#1 h Oct 26, 2005, 2:32 AM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ , $K$ and $L$ are points on the side $BC$ ( $K$ being closer to $B$ than $L$ ) such that $BC \cdot KL = BK \cdot CL$ and $AL$ bisects $\angle KAC$ . Show that $AL \perp AB.$

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frt

#2 h Oct 26, 2005, 3:58 PM • 6 Y

Y by Mathalete01, tantheta67, KKT, dumbduck, Adventure10, Mango247

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problemsolve_AoPS

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problemsolve_AoPS

#3 h Jun 10, 2016, 6:17 PM • 4 Y

Y by Rushil_india, Adventure10, Mango247, namanrobin08

Let $\angle BAK = \gamma, \angle KAL = \angle LAC = \alpha, \angle ACB = \beta$ .
We know $BC \cdot KL = BK \cdot CL$

By Sine Rule,
$\frac{KL}{sin \alpha}=\frac{AK}{sin(\alpha+\beta)}$ and $\frac{CL}{sin \alpha}=\frac{AC}{sin(\alpha+\beta)}$
Again, $\frac{BK}{sin \gamma}=\frac{AK}{sin(2\alpha+\beta+\gamma)}$ and $\frac{BC}{sin 2\alpha+\gamma}=\frac{AC}{sin(2\alpha+\beta+\gamma)}$

Equating these, we get $\frac{KL}{CL}=\frac{AK}{AC}=\frac{\frac{BK}{sin \gamma}}{\frac{BC}{sin (2\alpha+\gamma)}}$

Note that we know $BC \cdot KL = BK \cdot CL \implies \gamma = 180-2\alpha-\gamma$ as $\gamma \neq 2\alpha+\gamma$ .
Thus we get $\gamma=90^o-\alpha$
$\angle BAL = \gamma + \alpha = (90^o - \alpha)+\alpha = 90^o \implies AL \perp AB$ .

Done

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634 posts

Reynan

#4 h Jun 10, 2016, 6:49 PM • 3 Y

Y by Pluto1708, Adventure10, Mango247

$\mathcal{H}(B,L;K,C)$ and $AL$ bisects $\angle KAC\implies AL\perp AB$

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1915 posts

#6 h Sep 8, 2019, 7:58 PM • 1 Y

Y by Adventure10

$BC.KL=BK.CL\implies\frac{BC}{BK}=\frac{CL}{KL}=\frac{AC}{AK}\implies\frac{AC}{BC}=\frac{AK}{BK}$ . Let $Z$ be a point on $AB$ such thtat $KZ$ is the angular bisector of $\angle AKB$ . So, $\frac{AK}{BK}=\frac{AZ}{BZ}=\frac{AC}{BC}$ . Therefore $CZ$ is the angular bisector of $\angle ACB$ . Now let $CZ\cap AL=M$ . Let $\angle KAL=\angle LAC=\theta$ . Let $\angle ACZ=\angle ZCB=\alpha$ . Let $\angle AKZ=\angle ZKB=\beta$ . Now $\angle KMC=90^\circ+\theta\implies\angle ZMK=90^\circ-\theta$ . Now, $2\theta+2\alpha=2\beta\implies\theta+\alpha=\beta\implies \angle LAC+\angle ACZ=\angle AMZ=\angle AKZ=\beta$ . Therefore $AMKZ$ is a cyclic quadrilateral. So, $\angle BAK=90^\circ-\theta\implies \angle BAL=90^\circ$ . So, $AL\perp AB$ .

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1915 posts

#13 h Oct 3, 2019, 5:35 PM • 2 Y

Y by Pakistan, Adventure10

Another Solution using Projective Geometry.

Solution: Let $l$ be a line $\perp$ to $AL$ passing through $l$ and let $l\cap AK=P$ and $l\cap AC=Q$ and let $AB\cap QP=O$

Now $-1=(C,K;B,L)\overset{A}{=} (Q,P;O,L)$ .

Now notice that $LQ=LP$ as $\triangle PAL\cong\triangle QAL$ . So, $\frac{OQ}{OP}=1\implies OP=OQ$ . So, $O$ must be the point at infinity. So, $AB\|PQ\implies AL\perp AB$ .

@below no. :no:

:no:

This post has been edited 1 time. Last edited by amar_04, Oct 3, 2019, 6:07 PM

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#14 h Oct 3, 2019, 5:51 PM • 1 Y

Y by Adventure10

Is Projective Geometry allowed in RMO

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bal09

#15 h May 29, 2024, 10:20 AM

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We can aplly stewart theorm

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#16 h Mar 5, 2025, 9:07 AM

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I did it using External Angle Bisector Theorem
By Internal Angle Bisector Theorem in CAK,
CA/AK = CL/LK , Also, BC/BK = CL/LK (Given)
-> CA/AK = BC/ BK , Extend CA to a point D
By Converse of External Angle Bisector Theorem in CAK,
AB is the bisector of external angle KAD .
Then using linear pair axiom angle KAB + KAL = 90
LAB = 90
HP.

This post has been edited 2 times. Last edited by oeliconeo, Mar 6, 2025, 8:43 AM
Reason: ...

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