Consider the triangle $\Delta ABC$. Suppose that $\vert \angle ABC\vert > \vert \angle BCA\vert$. Prove that $\vert AC\vert > \vert AB\vert$.
This question was a simple question and it was out of 20 marks and I got 9/20 what have I done wrong? I must use Euclids Elements without trigonomety
Suppose for the sake of contradiction that if $\vert \angle ABC \vert > \vert \angle BCA\vert$ then $\vert AC\vert \leq \vert AB\vert$. Case 1: Suppose $\vert AC\vert = \vert AB\vert$, then we would have an isosceles triangle and by (P.1.3) we would have $\vert \angle ABC \vert=\vert \angle BCA\vert$ which is false. Case 2: Suppose $\vert AC \vert < \vert AB \vert$ then by (E.I.18) we would have $\vert \angle ABC \vert < \vert \angle BCA \vert$ which is false. Therefore by contradiction we have shown that the statement if $\vert \angle ABC \vert > \vert \angle BCA\vert$ then $\vert AC\vert \leq \vert AB\vert$ is false and thus it follows that the original proposition is true.
SoluTion: Draw
on AC point D, that <DBC = <ACB
Therefore: |BD| = |CD|
By triangle inequality: |DA| + |BD| > |AB|
We know that |BD| = |CD|
Therefore
|AC| = |CD| + |DA| = |BD| + |DA|
And we know that |DA| + |BD| > |AB|
Therefore |AC| > |AB|.