 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Hardest easy geometry problem (Posted on 2016-11-04) Given: triangle ABC with D on AC and E on BC,
angle CAE = 10°,
angle BAE = 70°,
angle CBD = 20°, and
angle ABD = 60°.

Find angle AED using only elementary geometry (no sines, cosines , or any other trigonometry.)

 No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Thoughts | Comment 1 of 2

`This is a fantastic problem!!!`
`In the following, three capital letters denote`
`an angle.`
`   I used the sine rule multiple times to derive`
`a formula for AED in terms of arbitrary values`
`for ABD, BAE, CAE, and CBD.`
`   It gave the correct values for AED in`
`Geometer's Sketchpad most of the times`
`(I'm still working on the other times).`
`   It gave AED = 20.00000 when I specified`
`the four values of our problem. `
`   For our problem without using trig. I could`
`find AED = 20 (with proof) only if I could`
`prove the following lemma:`
`           The foot of the perpendicular from`
`        point P to side BC is the midpoint of`
`        line segment D'E.`
`           Point P is the intersection of line`
`        segments AE and D'E'. Points D' and E'`
`        are the reflections of points D and E`
`        about the bisector of ACB.`
`   I'm still trying to prove the lemma. It seems`
`to hold in Geometer's Sketchpad. Any hints`
`or help would be appreciated.`
`   One point I would like to make: I would not even`
`try this problem without  Geometer's Sketchpad.`

 Posted by Bractals on 2016-11-16 15:31:08 Please log in:
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