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Let $ABC$ be an acute triangle and $\omega$ its circumcircle. The bisector of $\angle ABC$ intersects $AC$ at $B_1$ and $\omega$ at $M$. An altitude from $B_1$ to $BC$ meets $\omega$ at $K$ and an altitude from $B$ to $AK$ meets $AC$ at $L$.Proove that $M,L,K$ are collinear.

I think angle chasing should be enough. I tried to poove that $\angle AKL = \frac{\angle B}{2}$. Then lots of quadrilateral should be cyclic. Unfortunately , I didn't find a solution. Please help me.

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  • $\begingroup$ Please add a diagram and give details of your attempt. As it stands, this question is likely to get downvoted and / or closed. $\endgroup$
    – Math Lover
    Commented Nov 18, 2021 at 6:42
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    $\begingroup$ Since this is tagged contest-math, could you please add a source, so that we know it’s not from an ongoing contest? $\endgroup$
    – Aphelli
    Commented Nov 18, 2021 at 8:44

1 Answer 1

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Collinearity

To prove $\measuredangle AKL = \dfrac{\measuredangle ABC}{2}$ (see in $\mathrm{Fig.\space 2}$), we need the following simple lemma.

$\bf{Lemma}:$

As shown in $\mathrm{Fig.\space 1}$, Points $E$ and $F$ are taken on the extended sides $AD$ and $BC$ respectively of the cyclic quadrilateral $ABCD$, such that its side $CD$ is parallel to the segment $EF$. Show that the quadrilateral $ABEF$ is cyclic.

Before reading our proof hidden in the spoiler box below, we would like to see OP trying to prove it him/her-self.

$\bf{Proof\enspace of\enspace Lemma}:$

Let $\measuredangle ABF = \psi$. Since the external angle at a vertex of a cyclic quadrilateral is equal to the internal angle at its opposite vertex, we have, $$\measuredangle EDC = \measuredangle ABF = \psi. \tag{1}$$ Furthermore, Since $CD$ is parallel to $EF$, we shall write, $$\measuredangle AEF = \measuredangle EDC = \psi. \tag{2}$$ It is evident from (1) and (2) that $\measuredangle AEF= \measuredangle ABF$. This means that the straight line $AF$ subtends equal angle at $B$ and $E$. According to the converse of the theorem Euclid-III-21, $ABEF$ is a cyclic quadrilateral.

$\mathbf{Proof\enspace of\enspace Parallelity \enspace of} \enspace \pmb{AC}\enspace \mathbf{and}\enspace \pmb{PQ} :$

As shown in $\mathrm{Fig.\space 2}$, segments $AK$ and $BL$ intersect with each other at $P$, while segments $B_1K$ and $BC$ at $Q$. We add the line segments $BK$, $MA$, and $PQ$ to the original configuration described by OP in the problem statement. Furthermore, for brevity, let $\measuredangle ABC = 2\phi$, $\measuredangle BCA = \alpha$, and $\measuredangle LBC= \beta$.

Since $\measuredangle BKA$ and $\measuredangle BCA$ are in the same segment of the circle $\omega$, we have, $$\measuredangle BKA = \measuredangle BCA = \alpha. \tag{3}$$

Since $\measuredangle KPB =\measuredangle KQB = 90^o$, according to the converse of the theorem Euclid-III-21, the quadrilateral $BKQP$ is cyclic. This means that $\measuredangle BKP$ and $\measuredangle BQP$ are in the same segment of the circle $BKQP$. Therefore, using (3), we shall state, $$ \measuredangle BQP = \measuredangle BKA =\alpha.$$

This means that $\measuredangle BQP = \measuredangle BCA$, and, hence, $PQ$ is parallel to $CA$.

$\mathbf{Proof\enspace of}\enspace \pmb{\measuredangle AKL = \dfrac{\measuredangle ABC}{2}} :$

The points $B_1$ and $L$ lies on the extended sides $KQ$ and $BP$ respectively of the cyclic quadrilateral $BKQP$. According to the lemma proven above, $BKLB_1$ is also a cyclic quadrilateral.

Since $BB_1$ is the angle bisector of $\measuredangle ABC$, we have $\measuredangle B_1BC = \phi$. With this we can determine $\measuredangle B_1BL$ as shown below. $$\measuredangle B_1BL = \measuredangle B_1BC - \measuredangle LBC = \phi - \beta$$

Since $BKLB_1$ is a cyclic quadrilateral, according to the theorem Euclid-III-21, $$\measuredangle B_1KL = \measuredangle B_1BL = \phi - \beta. \tag{4}$$

In a similar vein, since $BKQP$ is a cyclic quadrilateral, according to the theorem Euclid-III-21, $$\measuredangle PKQ= \measuredangle PBQ = \beta. \tag{5}$$

Using (4) and (5), we can express the magnitude of the $\measuredangle AKL$ as $$\measuredangle AKL = \measuredangle PKQ + \measuredangle B_1KL = \phi. \tag{6}$$

We describe below how to prove that $K$, $L$, and $M$ are collinear. Here too, we encourage OP to work out own proof before reading ours.

$\mathbf{Proof\enspace of\enspace collinearity\enspace of\enspace} \pmb{K},\enspace \pmb{L},\enspace \mathbf{ and }\enspace \pmb{M} :$

Since $\measuredangle AKM$ and $\measuredangle ABM$ are in the same segment of the circle $\omega$, we have, $$ \measuredangle AKM = \measuredangle ABM = \phi. \tag{7}$$ It follows from (6) and (7) that $$\measuredangle AKM = \measuredangle AKL.$$ This can be possible if and only if $K$, $L$, and $M$ are collinear.

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