Question

in the given figure ABCD and PQRC are rectangle and Q is the midpoint of AC prove that
1)DP=PC
2) PR=1/2AC

Answer 1

Heyah!


Given that PQRS and ABCD are rectangle and Q is the mid pt of AC.

Since PQRS and ABCD are rectangle, every angle would be of 90°.

Consider AD and PQ.

angle ADP = 90°
angle QPC = 90°

Now, angle ADP lies corresponding to angle QPC and are equal.

If corresponding angles are equal, then the lines are parallel


=> AD || PQ

Now consider ∆ADC.

Q is the midpoint of AC (Given)
AD || PQ (shown above)

We know that,
line drawn parallel from the midpoint of a side, to other side, bisects the third side.

Hence, P would be the midpoint of DC

Hence, DP = PC

2)

To prove
PR = 1/2 AC

In a rectangle, diagonals are equal.

=> In PQRS, PR = QC.

Now, Q is mid pt of AC

=> QC = 1/2 AC

and QC = PR

=> PR = 1/2 AC



Hence Proved :)

Answer 2

here is your answer OK ☺☺☺☺☺☺☺☺


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]