prove: there cannot be a triangle in which the trisectors of an angel also trisect the opposite side?

Question

i've been trying and trying to figure this out for 2 days! can someone give me an indirect proof?

Answer 1

The altitude of a triangle is a line segment extending from any vertex of a triangle perpendicular to the line containing the opposite side.


also... an isosceles triangle is a triangle with two congruent sides.

Answer 2

Okay, take a triangle (take a 30-60-90 right triangle to get a view of what I'm talking about).
Trisect the 60º angle.

we must assume that the three partitions on the leg opposite the 60º angle are equal. We can define the side adjacent to the 60º angle to be B. The partitions all are length A.
Doing some nifty triangle work, we come to the following equations:

A = B / tan 70
2A = B / tan 50
3A = B / tan 30

or
A tan 70 = 2A tan 50 = 3A tan 30.
Now prove they aren't equal (calculator exercise).

THEN, you can use the "any triangle" identities (ab cos Θ) to get equalities in A, and then prove they aren't equal in a random triangle. More work but possible.

Answer 3

Try to prove that there is one. Try to prove that two lines radiating from a common angle, will cut the opposite side in 3 equal segments.
Then try to show what kind of triangle (isoceles, right angle) is needed to fulfill the condition.

If you are lucky, you will quickly arrive at a contradiction, or, if less lucky, you may have to exhaust all possible forms of triangles.