Geometry Proof/Trisect angles and sides...?

Question

Prove that it is impossible to have a triangle in which the trisectors of an angle also trisect the opposite side.

Answer 1

let ABC be triangle >> A+B+C=180

divide A into (3 times ) (A/3)

side BC > will have 2 points D & E

BD=DE=EC = a/3

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join AE, AD

AE is median for triangle CAD, extend AE to A'E such that AE=A'E

use congruence property with (A/3) angles on top >>> you will get

AC = b (side) = AD >>> isosceles

similarly

use construction for point D

AB = side (c) = AE

use exterior angle theorem

C = A/3 + B

B = A/3 + C

add

A = 0

impossible to have such a triangle

analyse yourself >