Question

show that the quadrilateral formed by
joining the med-point of the consecutive
side of a square is also a square.​

Answer 1

Answer:

Given:

P, Q, R and S+are themid-points+of AB, BC, CD and DA respectively.

To prove:

PQRS is a rhombus.

Proof:

Consider triangle BAC+

PQ|| AC and PQ+=+%281%2F2%29AC….(1)

(In a triangle the segment joining the mid-points of

two sides are parallel and equal to third side)

Consider triangle ADC+,

SR|| AC and SR+=+%281%2F2%29AC….(2)

From (1) and (2),

PQ|| SR and PQ=SR

PQRS is a parallelogram �����….(3)

AD+=+BC….. (opposite sides of a rectange)

So

%281%2F2%29AD+=%281%2F2%29BC…..

i.e. AS+=+BQ+

Consider triangle +APS+, and triangle +BPQ+,

AP+=+BP…. (P is the mid-point of AB)

AS+=+BQ

angle SAP = PBQ = 90degrees

So, triangle +APS+ is congruent to triangle +BPQ+.... (SAS congruency condition)

+PS+=+PQ ��������…..(4)

From (3) and (4),

PQRS is a parallelogram in which PS+=+PQ+.

PQRS is a rhombus .