Question

The diagonals of a quadrilateral intersecta
right angles. Prove that the figure obtained by
joining the mid-points of the adjacent sides of
the quadrilateral is a rectangle
​

Answer 1

Answer:

Let ABCD be a quadrilateral where P, Q, R, S are the midpoint of AB, BC, CD, DA. Diagonal AC and BD intersect at a right angle at point O. We need to show that PQRS rectangle.

Proof :

Here, ABCD is a quadrilateral.

AC and BD are diagonals, which are perpendicular to each other.

P,Q,R and S are the mid-point of AB,BC,CD and AD respectively.

In △ABC,

P and Q are mid points of AB and BC respectively.

∴ PQ ∥ AC and PQ= 1/2AC ----- ( 1 ) [ By mid-point theorem ]

Similarly, in △ACD,

R and S are mid-points of sides CD and AD respectively.

∴ SR ∥ AC and SR= 1/2AC ----- ( 2 ) [ By mid-point theorem ]

From ( 1 ) and ( 2 ), we get

PQ ∥ SR and PQ = SR

∴ PQRS is a parallelogram.

Now, RS ∥ AC and QR ∥ BD.

⇒ Also, AC ⊥ BD [ Given ]

∴ RS ⊥ QR

∴ PQRS is a rectangle.

Step-by-step explanation:

Hope it helps!