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Question no 2

Question no 2 Similarly, we can prove that MS NQ In the quadrilateral SMQN , since MQ = NS and MS NO, therefore EXERCISE II(A) I. ABCD is a parallelogram. From A and B perpendiculars AP and BQ are drawn to meet CD or CD produced. Prove that AP = BQ. 2. E and F are the mid-points of AB and AC tu.'0 sides of a AABC. P is any point on BC. AP cuts EF at Q. Prove that AQ = PQ. . E and F are the mid-points of sides AB and CD respectively of a parallelogram ABCD. Prove that AEFD is a parallelogram. ABCD is a parallelogram and its diagona intersect each other at O. Through O,

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Hey,
In triangle ABC, 
F is the midpoint of AB
E is the midpoint of AC
Therefore, 
FE//AB & FE=1/2AB(Midpoint theorem). 
Now, 
In triangle ABP,
F is the midpoint of AB
FQ//BP(FQ and BP are parts of FE and BC respectively) 
Therefore, 
Q is the midpoint of AP(Converse of midpoint theorem) 
Hence, AQ=QP.
Hence, proved. 
Hope this helps you!!!!

 
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