Using an indirect proof?

Question

USING AN INDIRECT PROOF: prove that there cannot be a triangle in which the trisectors of an angle also trisect the opposite side.

I do not know how to do this. Can someone show me how? And I mean show. Not just give me the answer because I want to learn.

Answer 1

In an indirect proof, we assume there can be such a triangle and show that the assumption leads to a conradiction.

Let's say we trisect angle C and extend the lines to opposite side AB, forming lines CD and CE (also forming lines AD, DE, and EB). The angle trisectors form triangles ACD, CDE, and BCE. Let's say that line AD = line DE = line EB, and show it is nonsense.

First I compute the two angles at point D. Triangles ACD and CDE have equal angles C/3, and equal sides opposite, AD and DE. They have a common side, CD, so the angles opposite that side must be equal to each other, by the law of sines, so angle A equals angle DEC. That means that the two angles at point D must be equal to each other, each 90 degrees.

By the same reasoning, the two angles at point E must be equal to each other, also each 90 degrees. Thus, triangle CDE contains two angles each of 90 degrees, plus one angle of C/3, which provides the contradiction in the original assumptions.

sahsjing's assertion about incongruous triangles is not correct. If you trisect the 90-degree angle in a 30-60-90 triangle, two adjacent new triangles formed will be congruent, but of course not all three. The two new, congruent, triangles will also be 30-60-90, and thus similar triangles to the original.

Answer 2

The hardest part in an indirect proof is to find contradiction.

For this problem, you assume these IS such a triangle that its angle trisectors trisect the opposite side.

Since between any two incongruous neighbor triangles you can find one common side and congruent bases but the different third sides. Then, by the cosine law, the base cannot be the same. Therefore, you find contradiction.

Answer 3

think of ~( if A then ~E ). Then A and E Then E or F Then D [through fact if (E or F) then D] Then C or D Then ~(A or B) [through fact ~(A or B) or ~(C or D)] Then ~A and ~B Then ~A Then if A then ~E A contradiction! the information is performed.