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.

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

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.