# In the given figure, ABCD is a square, M is the midpoint of AB and PQ ⊥ CM meets AD at P and CB produced at Q. prove that (i) PA=BQ, (ii) CP=AB+PA. 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

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