Question

given: △abc is not isos - celes; ap does not equal ac; pc does not equal ac; ∠1 ≅ ∠2
prove: △apc is not isos - celes (using the indirect method).

Explanation:

Step1: Assume the opposite

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

Step2: Analyze the cases

Case 1: If \(AP = AC\), this contradicts the given that \(AP\) does not equal \(AC\).
Case 2: If \(PC=AC\), this contradicts the given that \(PC\) does not equal \(AC\).
Case 3: If \(AP = PC\), then \(\angle PAC=\angle PCA\). But since \(\angle 1=\angle 2\), using the angle - side relationships in triangles (e.g., the Law of Sines or congruence - related concepts if we consider triangles \(ABP\) and \(CBP\) with \(BP = BP\) (common side), \(\angle 1=\angle 2\)), we would get a contradiction to the fact that \(\triangle ABC\) is not isosceles.

Answer:

1. Assume \(\triangle APC\) is isosceles.
2. A statement is either true or false.
3. If \(\triangle APC\) is isosceles, then either \(AP = AC\) or \(PC = AC\) or \(AP=PC\), but \(AP

eq AC\) (given), \(PC
eq AC\) (given), and if \(AP = PC\) we get a contradiction to \(\triangle ABC\) not being isosceles.