Answer:

Step-by-step explanation:

Given : -

ABCD is a parallelogram

P , Q , R & S are the mid points of the sides AB , BC , CD , DA

Required to PROVE : -

PQ = RS

QR = PS

Theorem used : -

Mid point theorem

This states that ;

The line segments joining the midpoints of the two sides of the triangle is parallel to the third side and also half of the third side .

Solution : -

ABCD is a parallelogram

P , Q , R & S are the mid points of the sides AB , BC , CD , DA

Here,

AC is a DIAGONAL .

Consider ∆ ABC & ∆ ADC

In ∆ ABC & ∆ ADC .

=> AC = AC ( side )

[ Reason : Common side ]

=> AB = CD ( side )

[ Reason : In a parallelogram, opposite sides are equal ]

=> AD = BC ( side )

[ Reason : In a parallelogram , opposite angles are equal ]

By using S.S.S. Congruency rule

∆ ABC  ∆ADC

However,

Similarly, if you want to write easily you can write that in a parallelogram a diagonal divides it into 2 Congruency triangles . so ∆ ABC  ∆ADC

This implies ;

PQ = ½ AC

RS = ½ AC

[ Reason : Mid point theorem ]

SINCE, RHS part is equal . Let's equate the LHS part .

Hence,

PQ = RS

Similarly,

Construction : -

Draw the diagonal BD in the ||gm PQRS .

Consider ∆ BCD & ∆ BAD

In ∆ BCD & ∆ BAD

=> BD = BD ( side )

[ Reason : Common side ]

=> BC = AD ( side )

[ Reason : Opposite sides are equal ]

=> CD = AB ( side )

[ Reason : Opposite sides are equal ]

By using S.S.S. Congruency rule ;

∆ BCD  ∆BAD

This implies ;

QR = ½ BD

QR = ½ BD PS = ½ BD

[ Reason : Mid point theorem ]

Since,

RHS part is equal . Let's equate the LHS part

Hence,

QR = PS

Hence Proved

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