Question

ABCD is a rhombus and P, Q, R, S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer 1

Hint- We will approach this question in such a way that we will consider the two triangles ABC and ADC respectively, deduce the relation between PQ and RS to prove PQRS is a parallelogram. After that we will use the property of rhombus that all sides of rhombus are equal to indirectly deduce the relation between $\angle 1,\angle 2$ . After that use SSS congruence rule in the triangle APS and CQR to make $\angle 3,\angle 4$ equal to each other. In the end, use the linear property of angles to deduce the relation between $\angle SPQ,\angle PQR$ and prove that $\angle SPQ$ is equal to ${90^0}$ and prove the already proven parallelogram to be a rectangle.

Complete step-by-step solution -

We have been given that ABCD is a rhombus where P, Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.
We have to prove that PQRS is a rectangle. For that we will first join A & C.
To prove that PQRS is a rectangle we will first prove it to be a parallelogram and then later prove one angle of it to be ${90^0}$
Let us consider $\Delta ABC$
Here we know that P and Q are the midpoints of sides AB and BC respectively.
We know that Line segments joining the mid-points of two sides of a triangle are parallel to the third side and are half of it.
$\therefore PQ\parallel AC$ and $PQ = \dfrac{1}{2}AC$ ----(1)
Let us consider $\Delta ADC$
Here we know that R and S are the midpoints of sides CD and AD respectively.
We know that the Line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
$\therefore RS\parallel AC$ and $RS = \dfrac{1}{2}AC$ ----(2)
From (1) and (2) we get,
$PQ\parallel RS $ and $PQ = RS$
Now in quadrilateral PQRS,
We can see that one pair of opposite sides is parallel and equal so we can say that PQRS is a parallelogram.
Now, we know that AB = BC because all sides of rhombus are equal.
So, we can also say that, $\dfrac{{AB}}{2} = \dfrac{{BC}}{2}$ and we know that P and Q are the midpoints of AB and BC respectively.
So, PB = BQ ----(3)
and AP = CQ ----(4)
Similarly, we know that AD = DC because all sides of rhombus are equal.
So, we can also say that, $\dfrac{{AD}}{2} = \dfrac{{DC}}{2}$ and we know that S and R are the midpoints of AD and DC respectively.
So, AS = CR ----(5)
and SD = DR ----(6)

Now in $\Delta BPQ$
PB = BQ (using (3))
$\therefore \angle 2 = \angle 1$ -----(7) (Angles opposite to equal sides are also equal)
Now in $\Delta APS\& \Delta CQR$
AP = CQ (Using (4))
AS = CR (Using (5))
PS = QR (Opposite sides of a parallelogram are equal)
$\therefore \Delta APS \cong \Delta CQR$ by SSS rule of congruence
$\angle 3 = \angle 4$ by CPCT ------(8)
Now, AB is a line
So, we can say that $\angle 3 + \angle SPQ + \angle 1 = {180^0}$ ----(9) as Linear pair angle sum $ = {180^0}$
Similarly, for line BC
$\angle 2 + \angle PQR + \angle 4 = {180^0}$ (Linear pair angle sum)
From (7) & (8) we get,
$\angle 1 + \angle PQR + \angle 3 = {180^0}$ --------(10)
From (9) and (10) we get,
$ \Rightarrow \angle 1 + \angle SPQ + \angle 3 = \angle 1 + \angle PQR + \angle 3$
$\therefore \angle SPQ = \angle PQR$ -----(11)

Now, $PS\parallel QR$ (Opposite sides of parallelogram are parallel)
& PQ is a transversal
So, $\angle SPQ + \angle PQR = {180^0}$ as interior angles on the same side of transversal are supplementary
Using (11) we get,
$ \Rightarrow \angle SPQ + \angle SPQ = {180^0}$
$ \Rightarrow 2\angle SPQ = {180^0}$
$ \Rightarrow \angle SPQ = \dfrac{{{{180}^0}}}{2} = {90^0}$
Now, PQRS is a parallelogram with one angle equals ${90^0}$ .
Hence, we can simply say that quadrilateral PQRS is a rectangle.
Hence showed.

Note- For such types of questions just keep in mind that the Line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. One must also be knowing that if one pair of opposite sides is parallel and equal, we can say that PQRS is a parallelogram. One must also be knowing about the SSS congruence rule of triangles and should also know that interior angles on the same side of transversal are supplementary.