isosceles wanted, line tangent to circumcircle, projection of orthocenter relate

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isosceles wanted, line tangent to circumcircle, projection of orthocenter relate

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Isosceles Triangle

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Source: Almaty Olympiad 2014 p1 , Kazakhstan

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#1 h Sep 27, 2018, 9:26 PM • 1 Y

Y by Adventure10

The line $l$ is the tangent to the circle circumscribed around the acute-angled triangle $ABC$ , drawn at the point $B$ . The point $K$ is the projection of the orthocenter of the triangle onto the line $l$ , and the point $L$ is the midpoint of the side $AC$ . Prove that the triangle $BKL$ is isosceles.

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#2 h Sep 28, 2018, 10:29 AM • 2 Y

Y by Adventure10, Mango247

Let $H$ be the orthocentre of $\triangle ABC$ , and let $D,F$ be the feet of the altitudes through $A,C$ , respectively. We will show that the triangles $DKL$ and $FBL$ are congruent.
We have $DL=FL$ , since both $D$ and $F$ lie on the Thales circle with diameter $AC$ .
$F,D,K$ clearly lie on the Thales circle with diameter $BH$ , and we have $\angle KBD = \angle BAC = \angle FDB$ . Thus the chords $KD$ and $BF$ have the same length.
We are left to show that $\angle LDK = \angle BFL$ ; this is easy angle chasing.
From these three facts we can conclude that $\triangle DKL$ and $\triangle FBL$ are congruent, thus $BL=KL$ .

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#3 h Jun 12, 2019, 8:50 AM • 2 Y

Y by Adventure10, Mango247

Let $H_A,H_C$ be foot from $A,C$ in $\Delta ABC$ , now since, $LH_C=LH_A$ and $BKH_CH_A$ is isosceles trapezium $\implies$ $LB=LK$

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