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Question Number 185729 by Mingma last updated on 26/Jan/23

Commented by HeferH last updated on 26/Jan/23

$$notclearwhatsegmentsarea,b,c$$

Commented by Mingma last updated on 26/Jan/23

The median BE of triangle ABC is trisected by its incircle. Calculate a:b:c

Commented by JDamian last updated on 26/Jan/23

I always was told that a triangle side is named using a lowercase letter after the uppercase vertex name located in front of that side -- the vertex having no contact with that side.

Commented by HeferH last updated on 26/Jan/23

$$Thatisafunnyrule,anyway,thankyoufor$$$$theclarification$$

Commented by Frix last updated on 26/Jan/23

$$\mathrm{Not}\mathrm{so}\mathrm{funny}.$$$$\alpha =\measuredangle BAC$$$$\mathrm{If}\mathrm{we}\mathrm{call}\mathrm{the}\mathrm{opposite}\mathrm{side}a\mathrm{etc}.\mathrm{we}\mathrm{get}$$$$\frac{\sin \alpha }{a}=\frac{\sin \beta }{b}=\frac{\sin \gamma }{c}$$$$\mathrm{and}$$$$a^2+b^2-2ab\cos \gamma =c^2$$$$\mathrm{where}\gamma \mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{of}a\mathrm{and}b$$

Commented by HeferH last updated on 27/Jan/23

$$I{don}^{\prime }tknowifsomeone{don}^{\prime }tusea,bandcfor$$$$verticesA,BandCgets-1pointorsomething$$

Commented by Frix last updated on 27/Jan/23

$${\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{been}\mathrm{out}\mathrm{of}\mathrm{all}\mathrm{institutions}\mathrm{where}\mathrm{they}$$$$\mathrm{can}\mathrm{take}\mathrm{points}\mathrm{from}\mathrm{you}\mathrm{for}\mathrm{more}\mathrm{than}15$$$$\mathrm{years}...$$$$\mathrm{You}\mathrm{can}\mathrm{use}\mathrm{whatever}\mathrm{you}\mathrm{want}\mathrm{but}\mathrm{you}$$$$\mathrm{have}\mathrm{to}\mathrm{explain}...$$$$\text{Prime causes double exponent: use braces to clarify}$$$$\mathrm{not}\mathrm{be}\mathrm{clear}\mathrm{enough}$$

Commented by HeferH last updated on 27/Jan/23

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Answered by mr W last updated on 27/Jan/23

Commented by mr W last updated on 27/Jan/23

$$sayAK=AH=x$$$$BF=FG=GE=m$$$${BD}^2=2m^2$$$${BK}^2=2m^2$$$${EH}^2=2m^2$$$$\Rightarrow BD=BK=EH=\frac{b}{2}-x$$$$AB=c=AE=\frac{b}{2}\Rightarrow \frac{b}{c}=2$$$$BC=a=\frac{b}{2}-x+b-x=\frac{3b}{2}-2x$$$$BE=3m=\frac{3\times BD}{\sqrt{2}}=\frac{3}{\sqrt{2}}(\frac{b}{2}-x)$$$$4\times {BE}^2=2a^2+2c^2-b^2$$$$18{(\frac{b}{2}-x)}^2=2{(\frac{3b}{2}-2x)}^2+2\times \frac{b^2}{4}-b^2$$$$20x^2-12bx+b^2=0$$$$\Rightarrow x=\frac{b}{10}$$$$\Rightarrow a=\frac{3b}{2}-2\times \frac{b}{10}=\frac{13b}{10}\Rightarrow \frac{a}{b}=\frac{13}{10}$$$$\Rightarrow a:b:c=13:10:5✓$$

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