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. So, either \(AP = AC\) or \(PC=AC\) (by the definition of an isosceles triangle).

Step2: Contradict the given

But we are given that \(AP\) does not equal \(AC\) and \(PC\) does not equal \(AC\). This contradicts our assumption.

Step3: Conclude

Since the assumption that \(\triangle APC\) is isosceles leads to a contradiction, the assumption is false.

Answer:

1. Assume \(\triangle APC\) is isosceles.
2. Definition of isosceles triangle.
3. If \(\angle1=\angle2\), \(BP = BP\) (common side), and if \(AP = PC\) (from the wrong assumption of isosceles \(\triangle APC\)), then by \(SAS\) (Side - Angle - Side) congruence criterion.
4. \(SAS\) (Side - Angle - Side) congruence.
5. \(AP = PC\) (from the wrong assumption of isosceles \(\triangle APC\) and using \(CPCTC\) on \(\triangle BPA\cong\triangle BPC\)) which would make \(\triangle ABC\) isosceles (if \(AP = PC\) and with other relations) but \(\triangle ABC\) is given non - isosceles. So, \(\triangle APC\) is not isosceles.