Given: ABCD is a square, AM = MB and PQ ⊥ CM

To prove: PA=BQ and CP=AB+PA

Proof:

In ∆AMP and ∆BMQ, we have

∠AMP = BMQ …vertically opposite angle

∠PAM = ∠MBQ = 90° …as ABCD is square

AM = MB …given

Thus by AAS property of congruence,

∆AMP ≅ ∆BMQ

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PA = BQ and MP = MQ …(1)

Now in ∆PCM and ∆QCM

PM = QM … from 1

∠PMC = ∠QMC … given

CM = CM … common side

Thus by AAS property of congruence,

∆PCM ≅ ∆QCM

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PC = QC

PC = QB + CB

PC = AB + PA …as AB = CB and PA = QB