Question

Point M lies on side AC of an equilateral triangle ABC; circles   \Omega _1  and   \Omega _2  are circumcircles of triangles MAB and MBC respectively. It is known that point   A  divides arc   MAB  of circle in the ratio of   \Omega _1  that MA:AB = 28:41. Find the ratio on which point C devides arc   MCB  of circle   \Omega _2 . In the answer indicate MC:CB

Answer 1

given

According to problem, figure is

Since ∇ABC

∇ABC is equilateral triangle, so AB=BC=CA=a

AB=BC=CA=a

Given that

$MA:AB=28:41$

We can write 

$MA:AC=28:41$

$⇒\frac{MA}{AC}=\frac{28}{41}$

Since

$MA=AC−MC$

So

$\frac{AC−MC}{AC}=\frac{28}{41}$

$\\⇒1−\frac{MC}{AC}=\frac{28}{41}$

$\\⇒\frac{MC}{AC}=1−\frac{28}{41}$

$\\⇒\frac{MC}{AC}=\frac{13}{41}$

Since $AC=CB$

$AC=CB$

So

$MC:CB=13:41$