 Hello and welcome to the session. I am Arsha and I am going to help you with the following problem which says show that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square. If we seated with a required quadrilateral such that it is diagonal BD and AC are equal and they bisect each other at right angles so we have to show that A, B, C, D is a square. Let us write down what we are given in a quadrilateral A, B, C, D in which diagonals AC and BD are equal also AC is perpendicular on BD and is equal to OC and OD is equal to OB and we have to prove that A, B, C, D is a square. Let us now begin with the proof triangle AOD triangle AOB AO is equal to AO which is common to both the triangles also OB is equal to angle AOD is equal to 90 degree that is both these angles are of 90 degree since AC is perpendicular on BD or we can say that AO is perpendicular on BD and also DO is equal to OB this is given to us that diagonals bisect each other so by this condition OD is concurrent to triangle AOB this implies AD is equal to AB since by CPCT that is corresponding parts of concurrent triangles are equal so let this be equation number one similarly we can show that triangle AOB is concurrent to triangle BOC and this will imply that site AB is equal to BC let this be equation number two also triangle BOC is concurrent to triangle DOC in the same way so this implies that DC is equal to BC let this be equation number three and lastly in the same manner we can show that triangle DOC is concurrent to triangle AOD which will further imply that AD is equal to DC let this be equation number four so from all these four equations we get that AB is equal to BC is equal to CD is equal to DA as all the four sides of correlator ABCD are equal now we will show that all these four angles are of 90 degree so first let us consider triangles ABD triangle BAC so in these two triangles site AD is equal to BC we have just proved above equal to BA this side is common to both the triangles and also we are given that the diagonals of correlator are equal so this implies BD is equal to AC and therefore by SSS congruence condition triangle ABD is congruent to triangle BAC which implies that angle DAB is equal to angle CBA this is by CPCT and now since both the pairs of 3D each other therefore the lateral CD is a parallelogram therefore we can say that angle DAB plus angle CBA is equal to 180 degree since AD is parallel to BC since we have proved above that ABCD is a parallelogram and AB is a transversal which intersects them then the sum of consecutive interior angles is supplementary now these two angles are equal just we have proved above so this implies two times of angle DAB is equal to 180 degree which further implies that angle DAB is equal to 180 degree upon 2 which is equal to 90 degree therefore angle DAB is equal to 90 degree and angle DAB is equal to angle CBA so both these angles are of 90 degree now again DC is parallel to AB AD is a transversal inside of a transversal as supplementary that is the sum of consecutive angles is supplementary angle CDA plus angle DAB is equal to 180 degree so this implies angle CDA is equal to 180 degree minus angle DAB it is equal to 180 degree minus angle DAB is 90 degree so this implies angle CDA is equal to 90 degree and similarly we can show that angle BCD is also equal to 90 degree so thus we have in quadrilateral ABCD equal degree each so this implies that quadrilateral CD is a so this completes the session take care and have a good day