Question

In the given figure, ABCD is a square. M is the
mid-point of AB and PQ is perpendicular to CM. CB
produced meets PQ at point Q.
Prove that :
(i) PA = BQ
(ii) CP = AB + PA.





pls answers only, otherwise i will report!!​

Answer 1

Answer:

ΔPAMandΔBMQ

∠PMA=∠BMQ=α

AM=MB

∠PAM=∠MBQ=90

∘

InΔCPQ

LM⊥PQ

PM=MQ

CP=CQ(isoscelesΔCPQ)

CQ=CB+BQ

CQ=AB+BQ

So,BA=BQ