It's not completely inscribed!

Geometry Level 4

Let $ABC$ be an equilateral triangle . Let $D$ be on $BC$ , $E$ and $F$ be on $AB$ and $G$ be outside $ABC$ such that $DG \mid \mid EF$ and $\angle DEF = \angle GFE = 90^{\circ}$ . Let $AC$ intersect $DG$ at $J$ and intersect $FG$ at $K$ . It is known $CD=25$ and $JK=16$ . Find the length of $EF$ .

The answer is 33.

1 solution

Victor Loh
Aug 16, 2016

Note that $\angle{KJG}=\angle{DJC}=\angle{BAC}=60^{\circ} \implies JG=16\cos{60^{\circ}}=8.$ Also $\angle{DJC}=\angle{JCD}=60^{\circ}$ implies $\triangle{DJC}$ is equilateral, so $EF=DG=DJ+JG=DC+JG=25+8=\boxed{33}$ as desired.

Perfect solution. +1

Sharky Kesa - 4 years, 10 months ago

Did the same way.

Niranjan Khanderia - 4 years, 9 months ago

Level 4 problem, 2 line solution. Seems legit. :P

Sharky Kesa - 4 years, 9 months ago

How got level 4

Vishwash Kumar ΓΞΩ - 4 years, 7 months ago

I dunno. If it gets popular, the level might go down.

Sharky Kesa - 4 years, 7 months ago

It's not completely inscribed!

The answer is 33.

1 solution

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