If P and Q are the middle points of the sides BC and CD of parallelogram ABCD respectively, prove that:
Attachments:
Answers
Answered by
5
Answer:
Since ABCD is a parallelogram,
D - A = C - B.
Rearranged, this says
B + D = A + C.
This will be useful for putting things in terms of A and C as need for the RHS.
Since P is the midpoint of BC,
P = (B+C)/2 => 2P = B + C
Similarly, since Q is the midpoint of CD,
2Q = C + D.
Then...
2(AP + AQ)
= 2( P - A + Q - A )
= 2P + 2Q - 4A
= B + C + C + D - 4A
= (B + D) + 2C - 4A
= A + C + 2C - 4A
= 3C - 3A
= 3( C - A )
= 3 AC
mdtaukir57700p5wcgg:
CAN u show me figure
I don't have a figure.
It's all vector algebra as shown above.
OK
IN LAST HOW U HAVE FINDED 3AC
Just like AP = P - A and AQ = Q - A, also AC = C - A. That's a general rule: The vector from X to Y is equal to Y - X.
ok and thank u once again
You're welcome. Glad to have helped!
Similar questions