Question

show that the quadrilateral abcd is a rectangle if the diagonals are equal and bisects each other. also in the quadrilateral mnop are the midpoint of the sides ab bc cd and da .show that the quadrilateral mnop is a rhombus​

Answer 1

Step-by-step explanation:

Given,

Diagonals are equal

AC=BD .......(1)

and the diagonals bisect each other at right angles

OA=OC;OB=OD ...... (2)

∠AOB= ∠BOC= ∠COD= ∠AOD= 90

0

..........(3)

Proof:

Consider △AOB and △COB

OA=OC ....[from (2)]

∠AOB= ∠COB

OB is the common side

Therefore,

△AOB≅ △COB

From SAS criteria, AB=CB

Similarly, we prove

△AOB≅ △DOA, so AB=AD

△BOC≅ △COD, so CB=DC

So, AB=AD=CB=DC ....(4)

So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is parallelogram

In △ABC and △DCB

AC=BD ...(from (1))

AB=DC ...(from $$(4)$$)

BC is the common side

△ABC≅ △DCB

So, from SSS criteria, ∠ABC= ∠DCB

Now,

AB∥CD,BC is the tansversal

∠B+∠C= 180

0

∠B+∠B= 180

0

∠B= 90

0

Hence, ABCD is a parallelogram with all sides equal and one angle is 90

0

So, ABCD is a square.

Hence proved.