Question

show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle

Answer 1

Given AC,BD are diagonals of a quadrilateral ABCD are perpendicular.

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

Proof:
In ΔABC, P and Q are mid points of AB and BC respectively.

∴ PQ|| AC and PQ = ½AC ..................(1) (Mid point theorem)

Similarly in ΔACD, R and S are mid points of sides CD and AD respectively.

∴ SR||AC and SR = ½AC ...............(2) (Mid point theorem)

From (1) and (2), we get

PQ||SR and PQ = SR

Hence, PQRS is parallelogram ( pair of opposite sides is parallel and equal)
Now, RS || AC and QR || BD.
Also, AC ⊥ BD (Given)
∴RS ⊥ QR.
Thus, PQRS is a rectangle.

Answer 2

Answer:

Given:- ABCD is a rhombus

E,F,G,H are the mid points of AB,BC,CD,DA

Formula used:- Line joining midpoints of 2 sides of triangle

Is parallel and half of 3rd side

Solution:-

BD is diagonal of rhombus

EH is the line joined by midpoints of triangle ABD,

∴EH is parallel and half of BD

GF is line joined by midpoints of side BC&BD of triangle BCD

∴GF is parallel and half BD

⇒ If HE is parallel to BD and BD is parallel to GF

∴ It gives HE is parallel to GF

⇒ If HE is half of BD and GF is also half of BD

∴ It gives HE is equal to GF

⇒ AC is another diagonal of rhombus

GH is the line joined by midpoints of triangle ADC,

∴GH is parallel to AC

∴GH is half of AC

FE is line joined by midpoints of side BC&AB of triangle ABC

∴FE is parallel to AC

∴ FE is half of AC

⇒ If GH is parallel to AC and AC is parallel to FE

∴ It gives GH is parallel to FE

⇒ If GH is half of AC and FE is also half of AC

∴ It gives GH is equal to FE

As diagonals of rhombus intersect at 90°

And AC is parallel to FE and GH

All angles of EFGH is 90°

All angles are 90° and opposite sides are equal and parallel

Conclusion:-

EFGH is a rectangle

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