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 $DF = BC$ .

#Geometry #Sharky

Easy Math Editor

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Comments

I am solving thia

Aditya Singh - 5 years, 6 months ago

Still solving?

Sharky Kesa - 5 years, 6 months ago

@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 - 5 years, 5 months ago

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 $\Delta FCD$ , $\dfrac{DF}{DC} = sin 30^{\circ} = \dfrac{2}{1}$ => $\dfrac{DF}{BC} = \dfrac{2}{2} = \dfrac{1}{1}$

Vishwash Kumar ΓΞΩ - 4 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$