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# Is there a pure geometric solution to this problem?

ABCD is a square. And there is a point E such that angle EAB = 15 and angle EBA = 15 degrees. Show that EDC is an equilateral triangle. Now there is a proof by contradiction solution to this problem. I was wondering if there is a pure geometric solution too(no trigonometry)?

Note by Shiv Gaur
3 years, 6 months ago

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Construction of a similar 15deg triangle on BC. Ie. BFC. Then BFE is equilateral. Then by congruency you can do it. Instead of proving CD=CE, prove that CB=CE. Sorry I'm slow with formatting so can't post entire solution. Hope you understand it.

- 3 years, 6 months ago

Proof by contradiction I meant: let angle EDC =x, angle DEC=(180-2x), angle EDA=(90-x), angle DEA=(x+15). Now suppose that x >60. Then 3x>180 therefore x>(180-2x). Therefore angle DCE>angle DEC. Therefore DE>DC. Also, x+15>75, so in triangle DEA we have angle DEA>angle DAE. Therefore AD>DE. Thus we have AD>DE>DC, an absurdity because AD=DC! Likewise for x<60. So the only option left is to conclude x=60 and this means triangle DEC is equilateral.

- 3 years, 6 months ago

x=arctan[2-tan15]

- 3 years, 6 months ago

Here is a geometric solution provided by Dr.Shailesh Shirali to me(along the lines suggested by Lavisha Parab):

Draw a copy of triangle EAB with DA as base. That is, locate point F inside the square such that triangle FDA is congruent to triangle EAB.

Then angle FAD = angle EAB = 15 deg, so angle FAE = 60 deg. Also, FA = EA. Hence triangle FAE is equilateral, and angle AFE = 60 deg, and FA = FD = FE.

Therefore F is equidistant from A, D, E and so is the circumcentre of triangle DAE.

Therefore by the circle theorems, angle AFE = 2 angle ADE.

But angle AFE = 60 deg. Hence angle ADE = 30 deg, and it follows that angle CDE = 60 deg. So triangle CDE is equilateral.

(I am unable to upload or paste the jpg of the above solution)

- 3 years, 6 months ago

You can also do it by area. Though I don't know whether that qualifies for "geometry".

- 3 years, 6 months ago

I see it as a pyramid (3D), just looking at the angles we cant really say it as equilateral. But an isosceles!

- 3 years, 6 months ago

I too searched a purely geometrical solution of it. I ended up using atleast the ratio of sides of a right triangle, which i guess not geometry... But let's see if someone comes out with a nice solution.... :)

- 3 years, 6 months ago