Suppose that ABCD is a Saccheri quadrilateral and that diagonals AC and BD intersect at point P. If M is the midpoint of DP and N is the midpoint of CP, prove that AM is congruent to BN.
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Proof. Suppose that ABCD is a Saccheri quadrilateral and that diagonals AC and BD intersect at point P. Assume M is the midpoint of DP and N is the midpoint of CP. We shall prove that AM is congruent to BN. By SAS, ∆ADB is congruent to ∆BCA. Therefore, PAB is congruent to PBA by CPCTC. Also, line segment PA is congruent to line segment PB by the Converse of the Isosceles Triangle Theorem. We can now say that ∆DAP is congruent to ∆CBP by SAS. Therefore, DAP is congruent to CBP by CPCTC and line segment CP is congruent to line segment DP by CPCTC. CP and DP are congruent. We are given that DM is congruent to MP and CN is congruent to NP by the definition of midpoint. Therefore, we can say that MP is congruent to NP by the transitive property. This allows us to say that ∆APM is congruent to ∆BPN by SAS. Therefore, AM is congruent to BN by CPCTC.

The part of my proof that worries me is the red underlined. Is it ok?