# Prove that these 3 points are in a straight line.

A triangle ABC is inscribed in a circle $\omega$. $BB_1$ bisects $\angle ABC$ (and so $M$ is the midpoint of the arc $AC$ ($B \notin AC$, where $AC$ is the arc)). $B_1K \perp BC$ ($K\in\omega$). $BL \perp AK$ ($L \in AC$). Prove that $K,L$ and $M$ are in a straight line.

I've used the letter D to represent those perpendicular angles in the picture (just so that they're easily seen). If you want to hear what I've tried, well, I can see that $\angle LBC = \angle AKB_1$, because $H_1H_2KB$ can be inscribed in another circle. I can't really tell anything more that could possibly be useful for solving this. I'm out of ideas and don't know where to start. Thanks.

• Why do you think $\;H_1H_2KB\;$ can be inscribed in a circle? I can't see, off the top of my head, that this quadrilateral has two opposite complementary angles, only two adyacent ones... – DonAntonio Dec 12 '13 at 17:57
• $\angle BH_1K = \angle BH_2K \implies H_1T \times TK = H_2T \times TB \implies H_1H_2KB$ can be inscribed in a circle. – user26486 Dec 12 '13 at 18:01
• I really don't understand why and how you deduce the above, but it never minds: a quadrilateral with two straight adjacent angles is a rectangle, and I don't think this one's a rectangle. – DonAntonio Dec 12 '13 at 18:18
• Well, I'm not going to show you all the proofs, you can simply find this way of finding out if a quadrilateral can be inscribed here: en.wikipedia.org/wiki/Cyclic_quadrilateral. It's written there that "If two lines, one containing segment $AC$ and the other containing segment $BD$, intersect at $X$, then the four points $A$, $B$, $C$, $D$ are concyclic if and only if $AX \times XC = BX \times XD$." – user26486 Dec 12 '13 at 18:31
• @DonAntonio: Here's a simpler explanation: $\angle BH_1K$ is a right angle, so $H_1$ is inscribed in the semi-circle with diameter $\overline{BK}$. Likewise for $H_2$. – Blue Dec 12 '13 at 22:41

The idea is to prove that $KL$ is the bisect of $\angle AKC$ so it should go through $M$. As you have said $\angle LBC = \angle AKB_1$. So it is enough to prove that $\angle B_1KL = \angle B_1BL$.

$\angle H_1=H_2=90 \land \angle BAK = \angle BCK \to \triangle ABH_1 \sim \triangle CKH_2 \to \frac{BH_1}{AH_1}=\frac{KH_2}{CH_2} \\ \angle H_1=H_2=90 \land \angle KAC = \angle KBC \to \triangle ALH_1 \sim \triangle BKH_2 \to \frac{H_1L}{AH_1}=\frac{KH_2}{BH_2}$

Divideing the equalities we will have $\frac{BH_1}{H_1L}=\frac{BH_2}{CH_2}$. So $H_1H_2||B_1L$, therefore we can write:

$\frac{H_1R}{H_2R}=\frac{RL}{RB_1}$

$\triangle BRH_2 \sim \triangle KRH_1 \to \frac{H_1R}{H_2R}=\frac{KR}{BR}$

$\frac{RL}{RB_1}=\frac{H_1R}{H_2R}=\frac{KR}{BR} \land \angle BRB_1=\angle KRL \to \triangle BRB_1\sim \triangle KRL$

Therefore $\angle RBB_1 =\angle RKL$ and this is what we wanted to prove.

• And thanks for the problem :-) – hhsaffar Dec 13 '13 at 21:58
• Wow. I've thought of an easier one. So we have to prove that $\angle BB_1K = \angle BLK$, which we can do by showing that $BB_1LK$ can be inscribed in a circle. And we can prove this by showing that $\angle LB_1K = \angle KBL$, which we know because $\angle BKA = \angle ACB$ (because $\triangle B_1H_2C$ and $\triangle BH_1K$ are right-angled). – user26486 Dec 14 '13 at 9:18
• @mathh Very good, that's really nice. – hhsaffar Dec 14 '13 at 11:48