show that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square

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  

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    (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  

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∠B+∠B= 180  

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∠B= 90  

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

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So, ABCD is a square.

Hence proved.

Step-by-step explanation:

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

answer is (b.

164 ft2)

a base  

18

b base  

7

h height  

8