Login

Answer the 14th (a)."DON'T SEND STHE SIMILAR QUERIES.KINDLY PERFORM THE STEP BY STEP AND APPROPRIATE ANSWER".:



14. (a) In figure ( 1) given below, A BCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram  ABPQ is completed. Prove that 
(i) the triangles ABX and QCX are congruent- 
(ii) DC=CQ=QP
 

Dear student

a)i) Consider △ABX and △QCX∠QXC=∠AXB      Vertically oppsote angles∠QCX=∠ABX     alternate interior anglesBX=XC        X is the mid point of BC⇒△ABX≅ △QCX    ASA criteriaii) Now since  △ABX≅ △QCX⇒AB=CQ    ...1     cpctAlso ABCD is a ∥gm⇒AB=CD   opposite sides are equal in a   ∥gm⇒CQ=CD   ....2   using 1Also ABPQ is a ∥gm⇒AB=PQ   opposite sides are equal in a   ∥gm⇒CQ=PQ      ....3  using 1So,  from 2 and 3 we haveCD=CQ=QPHence Proved.
Regards

  • 5
View Full Answer
(a) (i) angle AXB= angle CXQ (vertically opposite angles)
          angle ABX= angle XCQ (alternate angles)
         BX=XC (X is the mid point of BC)
        BY ASA congrence rule,
        THUS, triangle ABX is congruent to triangle QCX
​     
​
  • 1
What are you looking for?