Question

P, Q, R and S are the midpoint of the side AB, BC, CD and DA respectively of quadrilateral ABCD in which AC=BD. prove that PQRS is a rhombus ​

Answer 1

Answer:

Given, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Also, AC = BD and AC is perpendicular to BD.

In ΔADC, by mid-point theorem,

SR||AC and SR =

AC

In ΔABC, by mid-point theorem,

PQ||AC and PQ =

AC

PO||SR and PQ = SR =

AC

Now, in ΔABD, by mid-point theorem,

SP||BD and SP =

BD =

AC

In ΔBCD, by mid-point theorem,

RQ||BD and RQ =

BD =

AC

SP = RQ =

AC

PQ = SR = SP = RQ

Thus, all four sides are equal.

Now, in quadrilateral EOFR,

OE||FR, OF||ER

∠EOF = ∠ERF = 90

(Opposite angles of parallelogram)

∠QRS = 90

Hence, PQRS is a square.

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

${\huge\tt\bf{GIVEN}}\begin{cases}\sf{A\: quadrilateral\:ABCD}\\\sf{P,Q,R\:and\:S\:are\:midpoints\:of\:it's\:sides}\\ \sf{AC=BD} \end{cases}$

$\huge{\tt {TO\:PROVE :-}}$

• PQRS IS RHOMBUS
• So, PS=PQ=QR=RS (sides)

$\huge{\tt {PROOF :-}}$

In ∆ BCD,

By mid-pint theorem,

Q is the midpoint of BC and R is the midpoint of DC.

So, $\tt {BD||QR , QR = \dfrac {1}{2}BD}$.....(1)

In ∆ BAD,

By mid-pint theorem,

P is the midpoint of AB and S is the midpoint of AD.

So, $\tt {BD||PS , PS = \dfrac {1}{2}BD}$.....(2)

By (1) and (2),

$\tt {QR||PS , PS = QR = \dfrac {1}{2}BD}$.

Therefore, □ PQRS is parallelogram.

Now, In ∆ ABC,

By mid-pint theorem,

P is the midpoint of AB and Q is the midpoint of BC.

So, $\tt {PQ||AC , PQ = \dfrac {1}{2}AC}$.......(3)

Now, we know that AC=BD,

Therefore, (3) becomes,

$\tt {PQ||BD , PQ = \dfrac {1}{2}BD}$.......(4)

$\huge{\boxed {\tt {Therefore, PS=PQ=QR=RS}}}$

As, all sides are equal it is a rhombus.

Hence,proved!!!

$\rule{200}2$