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Question

# Let ABC be an acute-angled triangle triangle, and let D,E,F be points on BC,CA,AB respectively such that AD is the median, BE is the internal angle bisector and CF is the altitude. Suppose ∠FDE=∠C,∠DEF=∠Aand∠EFD=∠B. Prove that ABC is equilateral.

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Solution

## Since ΔBFC is right-angled at F, we have FD=BD=CD=a/2. Hence ∠BFD=∠B. Since ∠EFD=∠B, we have ∠AFE=π−2∠B. Since ∠DEF=∠A, we also get ∠CED=π−2∠B. Applying sine rule in ΔDEF, we haveDFsinA=FEsinC=DEsinB.Thus we get FE=c/2andDE=b/2. Sine rule in ΔCED givesDEsinC=CDsin(π−2B).Thus (b/sinC)=(a/2sinBcosB). Solving for cosB, we havecosB=asinc2bsinB=ac2b2.Similarly, sine rule in ΔAEF givesEFsinA=AEsin(π−2B).This gives (since AE=b/c(a+c)), as earlier,cosB=aa+c.Compairing the two values of cosB, we get 2b2=c(a+c). We also havec2+a2−b2=2cacosB=2a2ca+c.Thus4a2c=(a+c)(2c2+2a2−2b2)=(a+c)(2c2+2a2−c(a+c)).This reduces to 2a3−3a2c+c3=0. Thus (a−c)2(2a+c)=0. We conclude that a=c. Finally2b2=c(a+c)=2c2We thus get b=c and hence a=c=b. This shows that ΔABC is equilateral.

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