Answer to Question #247364 in Geometry for M.l

Question #247364

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


1
Expert's answer
2021-10-12T15:02:59-0400

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"

"\u21d2\\frac{MA}{AC}=\\frac{28}{41}"



Since


"MA=AC\u2212MC"


So

"\\frac{AC\u2212MC}{AC}=\\frac{28}{41}"


"\\\\\u21d21\u2212\\frac{MC}{AC}=\\frac{28}{41}"


"\\\\\u21d2\\frac{MC}{AC}=1\u2212\\frac{28}{41}"


"\\\\\u21d2\\frac{MC}{AC}=\\frac{13}{41}"


Since "AC=CB"

"AC=CB"

So

"MC:CB=13:41"




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