slicing problem with a right angle in the end

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slicing problem with a right angle in the end

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Source: Tuymaada 2007, Problem 6

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1145 posts

#1 h Jul 15, 2007, 8:24 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Point $D$ is chosen on the side $AB$ of triangle $ABC$ . Point $L$ inside the triangle $ABC$ is such that $BD=LD$ and $\angle LAB=\angle LCA=\angle DCB$ . It is known that $\angle ALD+\angle ABC=180^\circ$ . Prove that $\angle BLC=90^\circ$ .

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TomciO

#2 h Jul 15, 2007, 2:19 PM • 2 Y

Y by Adventure10, Mango247

Extend $AL$ to meet $BC$ at $E$ . Now $\angle DLE = 180^{\circ}-\angle ALD = \angle ABC$ . Since $\angle DLB = \angle DBL$ (triangle $DLB$ is isosceles) we must also have $\angle ELB = \angle EBL$ so the triangle $ELB$ is isosceles too. It means that $ED$ is a perpendicular bisector of $LB$ so $\angle LED = \frac{1}{2}\angle LEB = 90^{\circ}-\angle LBE$ . We also have $\angle LEC = 2\angle LBE$ so $\angle DEC = 90^{\circ}+\angle LBE$ . Since $\angle EAD = \angle DCE$ quadrilateral $ADEC$ is cyclic so $\angle DAC= 90^{\circ}-\angle LBE$ . Note that $\angle ACD = \angle LCE$ (from the conditions) and $\angle ADC = \angle AEC$ (because $ADEC$ is cyclic) so the triangles $ADC$ and $LEC$ are similar. Therefore $LEC = \angle DAC = 90^{\circ}-\angle LBE$ . So $\angle BLC = \angle BLE+\angle ELC = 90^{\circ}$ and we are done.

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Nemion

#3 h Jun 15, 2008, 11:26 PM • 2 Y

Y by Adventure10, Mango247

Let $AL$ meet $BC$ at $L_1$ , $\angle{DLL_1}=\angle{ABC}$ $\Rightarrow$ $LL_1=BL_1$ , quadrilateral $ADL_1C$ is cyclic, because $\angle{DAL_1}=\angle{L_1CD}$ then, $\angle{DL_1A}=\angle{DCA}$ $\Rightarrow$ $\angle{LCB}=\angle{DL_1A}=\angle{DL_1B}=90^o-\angle{LBL_1}$ then $\angle{BLC}=90^o$

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#4 h Jun 16, 2008, 4:40 AM • 2 Y

Y by Adventure10, Mango247

the main idea of this problem is prove ADIC is cyclic( many proof). Then <DIB=<A
so <ALI= 180-2<A. then <LCI = <A(1) ,but we have <LCB = 90-<DIB=90-<A(2)
From (1) and (2) we have <BLC= 90

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djisjustchilling

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djisjustchilling

#5 h Jun 16, 2008, 8:34 AM • 2 Y

Y by Adventure10, Mango247

Ya I was trying to follow the explanation but my geometry is not all up to date, is there like a online article or something I could read that would make the explanation more clear.

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#6 h Sep 25, 2015, 10:39 PM • 2 Y

Y by Adventure10, Mango247

Sorry to revive it... a nice one indeed,
Let $LD$ meet $CB$ again at $X$ . Then just note that our angle condition gives $A,L,B,X$ cyclic and moreover a trapezoid. Now its finished by some angle chase...

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