Q. 324.5( 8 Votes )

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.


Answer :

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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