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# The diagram has to be big! Big, I say!

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
1 year, 10 months ago

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Yes, It is very obvious.

$$[1]$$ In $$\Delta BEF$$, $$BC \perp EF$$ and $$EA \perp BF.$$ => $$D$$ is the orthocenter. => $$DF \perp BE$$.

$$[2]$$ Clearly, $$\angle CEB = 60^{\circ} = \angle CDF$$ => In right $$\Delta FCD$$ , $$\dfrac{DF}{DC} = sin 30^{\circ} = \dfrac{2}{1}$$ => $$\dfrac{DF}{BC} = \dfrac{2}{2} = \dfrac{1}{1}$$ · 5 months, 1 week 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. · 1 year, 9 months ago

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I am solving thia · 1 year, 9 months ago

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Still solving? · 1 year, 9 months ago

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