The 80-80-20 Triangle Problem, Solution #10
Let ABC be an isosceles triangle
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Copyright © 1996-2018 Alexander Bogomolny
The proof is due to Luke Rapley.
With radius CE and center C sweep an arc to intersect BC at U and AC at V.
ΔCEU is isosceles (CE = CU) and, ∠BCE being 60°, is also equilateral. In particular, EU = CE and
We are going to compare triangles BEU and VAE:
AE = EU, since both equal CE,
AV = BE, because AV = AC - CV = AB - AE = BE,
∠EAV = ∠BEU since both equal 20°.
We conclude that, by SAS, the two triangles are equal. From here we get
On the other hand, since ΔBCD is isosceles, BC = CD and consequently,
BU = CU - BC = CV - CD = DV.
Therefore, VD = EV and triangle DEV is isosceles. Now,
∠DVE = 180° - ∠AVE = 180° - ∠EBU = ∠ABC = 80°.
So ∠DEV = (180° - 80°)/2 = 50°. Counting angles around E, we see that ∠CED = 30°.
The 80-80-20 Triangle Problem
- Solution #1
- Solution #2
- Solution #3
- Solution #4
- Solution #5
- Solution #6
- Solution #7
- Solution #8
- Solution #9
- Solution #10
- Solution #11
- Solution #12
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Copyright © 1996-2018 Alexander Bogomolny
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