let us prove that
consider the parallelogram abcd with diagonals ac and bd as shown and they intersect at right angles at oa parallelogram is a square if its adjacent angles are 90° and adjacent sides are equalso we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a squareconsider δbad and δcdaac = bd … givenab = dc … opposite sides of a parallelogramad is the common sidetherefore, δbad ≅ δcda … sss test for congruency⇒ ∠bad = ∠cda …corresponding angles of congruent triangles …(i)as it is given that abcd is parallelogram⇒ ∠bad + ∠cda = 180° … sum of adjacent angles of a parallelogram is 180°using equation (i)⇒ ∠cda + ∠cda = 180°⇒ 2 × ∠cda = 180°⇒ ∠cda = 90°⇒ ∠bad = 90°thus, adjacent angles are 90° … (iii)consider δaob and δaod∠aob = ∠aod … both 90° because given that diagonals intersect at right anglesod = ob … diagonals of a parallelogram bisect each otherao is the common sidetherefore, δaob ≅ δaod … sas test for congruency⇒ ab = ad … corresponding sides of congruent trianglesthus, adjacent sides are equal … (iv)from statements (iii) and (iv) we can conclude that parallelogram abcd is a squaretherefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square.
given to prove 

then taking l.h.s.

we have 

                    





                                  

 

  

=  r.h.s.

hence proved. let me explain the... please check this video about. correct We will have this next week! Yep, we'll fix this soon. that is stupid Please post your code so that we can help you. congratulations to you