ABCD and PQRC are rectangles and Q us the mid-point of AC show that P is the mid-point of DC and pr is half of ac
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Given, ABCD and PQRC are rectangles in which Q is mid-point of AC.
To Prove: DP=PC and PR=1/2AC
Proof:
Given, PQRC is a rectangle which implies that PQ || RC ⇒ PQ || BC.
In triangle BCD, we have
Q mid point of BD and PQ || BC.
Therefore, P is mid-point of CD. [Using converse of mid-point theorem which states that the line drawn through the mid-point of one side of a triangle,parallel to another side, intersects the third side at its mid-point.]
⇒ DP = PC.
Again, in triangle BCD, P is mid-point of CD and PC || Qr which implies that DC || QR.
Again, R is mid-point of BC. [By converse of mid-point theorem]
Again, in triangle BCD, P is mid-point of CD and R is mid-point of BC.
Therefore, PR = 1/2 BD. [Using mid-point theorem which states that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it]
We know that both the diagonals of a rectangle are equal.
So, AC = BD.
⇒ PR = 1/2 AC
[Hence Proved]
To Prove: DP=PC and PR=1/2AC
Proof:
Given, PQRC is a rectangle which implies that PQ || RC ⇒ PQ || BC.
In triangle BCD, we have
Q mid point of BD and PQ || BC.
Therefore, P is mid-point of CD. [Using converse of mid-point theorem which states that the line drawn through the mid-point of one side of a triangle,parallel to another side, intersects the third side at its mid-point.]
⇒ DP = PC.
Again, in triangle BCD, P is mid-point of CD and PC || Qr which implies that DC || QR.
Again, R is mid-point of BC. [By converse of mid-point theorem]
Again, in triangle BCD, P is mid-point of CD and R is mid-point of BC.
Therefore, PR = 1/2 BD. [Using mid-point theorem which states that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it]
We know that both the diagonals of a rectangle are equal.
So, AC = BD.
⇒ PR = 1/2 AC
[Hence Proved]
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