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Let \(ABC\) be a triangle with \(\angle BAC=120^{\circ}\). Let \(D\) be the midpoint of \(BC\) and suppose that \(AD\) is perpendicular to \(AB\). Let \(E\) be the second point of intersection of the line \(AD\) with the circumcircle of triangle \(ABC\). Let \(F\) be the intersection of line \(AB\) and \(CE\).

(a) Prove that line \(DF\) is perpendicular to line \(BE\).

(b) Prove that \(DF = BC\).

Note by Sharky Kesa
10 months, 2 weeks ago

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@Sharky Kesa : I am able to prove only the part (a). My proof is : It can be observed that D is the orthocentre of triangle BFE. Hence the result follows. Shrihari B · 9 months, 1 week ago

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I am solving thia Aditya Singh · 10 months ago

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@Aditya Singh Still solving? Sharky Kesa · 9 months, 4 weeks ago

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