Question

if the diagonals of a parallelogram intersect at right angles ..prove that it is a rhombus .

Answer 1

A quadrilateral ABCD in which  the diagonals AC and BD intersect at O such that AO = OC , BO = OD and AC is perpendicular to BD.

To prove : ABCD is a rhombus.
First Proof : Since the diagonals AC and BD of quadrilateral ABCD bisect each other at right  angles.
Therefore,  AC is the perpendicular bisector of the segment BD.
A and C both are equidistant from B & D
So, AB = AD and CB = CD ... (1)
Also, BD is the perpendicular bisector of line segment  AC.
B and D both are equidistant from A & C.
Thus, AB = BC and AD = DC ... (2)
From (1) and (2), we get
AB = BC = CD = AD
Thus , ABCD is a quadrilateral whose diagonals bisect each other at right angles and all four sides are equal.
Hence, ABCD is a rhombus.
Second Proof : First we shall prove that ABCD is a || gm.
In ∆s AOD = COB,
i) AO = OC (given)
ii) OD = OB (given)
iii) Angle AOD = Angle COB (vertically opposite angles)
By SAS criterion, ∆AOD is congruent to ∆COB
=> Angle OAD = Angle OCB ...(i) [CPCT]
Now, line AC intersects AD and BC at A and C respectively such that
Angle OAD = Angle OCB (alternate angles are equal)
Therefore, AD || BC
Similarly, AB || CD
Hence , ABCD is a parallelogram.



Now, In ∆s AOD = COB,
i) AO = OC (given)
ii) OD = OD (common)
iii) Angle AOD = Angle COB (vertically opposite angles)
By SAS criterion, ∆AOD is congruent to ∆COB
=> AD = CD ....(ii) [CPCT]
Now, since ABCD is a parallelogram [Proved above]
Then, AB = CD and AD = BC [Opp. sides of a parallelogram are equal]
AB = CD = AD = BC [Using (ii)]
Hence, quadrilateral ABCD is a rhombus.

Answer 2

your answer is in attachment above.