Question

given: \\( \triangle abc \\) is not isos - celes; ap does not equal ac; pc does not equal ac; \\( \angle 1 \cong \angle 2 \\) prove: \\( \triangle apc \\) is not isos - celes (using the indirect method).

Explanation:

Step1: Assume the opposite

Assume that \(\triangle APC\) is isosceles.

Step2: Analyze the consequences

If \(\triangle APC\) is isosceles, then either \(AP = AC\) or \(PC=AC\) (by the definition of an isosceles triangle). But we are given that \(AP
eq AC\) and \(PC
eq AC\). This is a contradiction.

Answer:

The reason for \(\angle1\cong\angle2\) is given. For statement 5 (assuming we are following the indirect proof steps), the statement could be "Assume \(\triangle APC\) is isosceles" and the reason "Indirect proof assumption". Then the next statement (if we follow the contradiction) could be "Contradiction to given \(AP
eq AC\) and \(PC
eq AC\)" with the reason "Based on isosceles triangle property (two sides equal) and given conditions". But since the problem seems to focus on \(\angle1\cong\angle2\) (maybe as a given for the overall proof), if we just consider the line \(4\): The reason for \(\angle1\cong\angle2\) is "Given".