Angle bisectors and circumscribed circle

Question

AL and BM are bisectors in $\triangle ABC$. The second common point of the circles described (circumscribed) around $\triangle ACL$ and $\triangle BCM$ lies on the side AB. Find $\angle ACB$.

I am not sure about the word "described" in the problem and if I am not right, I would be very grateful if you correct me.

Look at the drawing to understand what I have noticed. $$\alpha + \beta = 90^\circ - \frac{\gamma}{2} $$ $$ \angle AFB = 90^\circ + \frac{\gamma}{2}$$

I am not sure that these statements will help with the solution but I just want to show I have tried sth.

This problem is screaming for an inversion with center at $C$. — user647486, Apr 19 '19 at 20:03
@user647486, I am 8th grade, and I don't know what inversion is. Also, I am not sure what you advise me to do in order to solve the problem. — Math Student, Apr 19 '19 at 20:07
Is it also given that $M$ and $L$ are centers of the circles? — Vasili, Apr 19 '19 at 20:16

user · Accepted Answer · 2019-04-20T07:40:14.413

1

Hint:

$$\begin {cases}\angle AEC=\angle ALC=\pi-\alpha-\gamma\\ \angle BEC=\angle BMC=\pi-\beta-\gamma\\ \alpha+\beta=\pi/2-\gamma/2 \end {cases}\\ \implies \pi=\angle AEC+\angle BEC=(\pi-\alpha-\gamma)+(\pi-\beta-\gamma)=\frac32 (\pi-\gamma). $$

Can you take it from here?

edited Apr 20 '19 at 07:40

answered Apr 19 '19 at 20:22

user

26,272

Thank you for your response! But why we use $\pi$ here? Does it signify degrees? – Math Student Apr 19 '19 at 20:27
@NikolDimitrova $\pi=180^\circ $. If you wish replace it everywhere. – user Apr 19 '19 at 20:43
As I see you used $\alpha, \beta, \gamma$ for the full angles, right? (not their halves as in my drawing)? – Math Student Apr 19 '19 at 20:45
@NikolDimitrova I have used $\alpha$ and $\beta $ for the half-angles as in your drawing, and $\gamma $ for the full angle $C $ as in your question. – user Apr 19 '19 at 20:48
@NikolDimitrova Because it is equal $\angle BMC $. – user Apr 19 '19 at 20:53
Okay, I got that but how to find $\angle ACB$ then? Sorry for the questions but I really want to understand it. I appreciate your help! – Math Student Apr 19 '19 at 20:56
@NikolDimitrova What kind of problem do you have with solving the last equation? – user Apr 19 '19 at 20:59
I can solve it ($\gamma = 60^\circ$), but I don't know where does it come from... – Math Student Apr 19 '19 at 21:02
@NikolDimitrova It comes from the solution of the equation. And you solved it correctly! – user Apr 19 '19 at 21:03
Nah, I don't understand how to write the equation after I had put the degrees on the drawing. DRAWING – Math Student Apr 19 '19 at 21:06
$\frac {3}{2} (\pi - \gamma)$ ? – Math Student Apr 19 '19 at 21:07
$(\pi-\alpha-\gamma)+(\pi-\beta-\gamma)$ ! – user Apr 19 '19 at 21:10
I understand that this sum must be equal to $180^\circ$. But from where did you get $\frac {3}{2} (\pi - \gamma)$? – Math Student Apr 20 '19 at 05:29
@NikolDimitrova From the equation $\alpha+\beta=\pi/2-\gamma/2$ which was derived already in your question. – user Apr 20 '19 at 07:18

Angle bisectors and circumscribed circle

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