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.)
This is a fantastic problem!!!
In the following, three capital letters denote
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