Langley’s Adventitious Angles

Geometry Level 2

A B C ABC is an isosceles triangle.
B = C = 8 0 ∘ B = C = 80 ^\circ .
C F CF at 3 0 ∘ 30^\circ to A C AC cuts A B AB in F . F. B E BE at 2 0 ∘ 20^\circ to A B AB cuts A C AC in E E .

Find then angle B E F BEF in degrees.

One of the all-time most famous geometry problems! Will wreck your brains if you haven't done this earlier, surely!


The answer is 30.

Aniruddha Bhattacharjee by Aniruddha Bhattacharjee , 25, India

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2 solutions

I am not giving the solution since one can find several solutions in Google.com.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Draw BG at 20 degrees to BC cutting CA in G.

Then angle GBF = 60 degrees and angles BGC and BCG are 80 degrees. So BC = BG.

Also angle BCF = angle BFC = 50 degrees, so BF = BG and triangle BFG is equilateral.

But angle GBE = 40 degrees = angle BEG, so BG = GE = GF.

And angle FGE = 40 degrees, hence GEF = 70 degrees and BEF = 30 degrees.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Simply elegant.

Greg Grapsas - 5 years, 2 months ago

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