if the diagonals are a parallelogram are equal , then show it is a rectangle
Answers
Answered by
2
Let ABCD be that parallelogram
Then AC=BD.
Now in ABD and BCD,
BD=BD
AB=CD
AD=BC
So, ABD is congruent to BCD
=> angle A = angle C
We know that the sum of opposite angles of a parallelogram is 180.
So, angle A + angle C= 180
=> 2angle A =180
So, angleA=90.
Since a parallelogram with opposite sides equal and all angles 90 degree, is a rectangle.
Then AC=BD.
Now in ABD and BCD,
BD=BD
AB=CD
AD=BC
So, ABD is congruent to BCD
=> angle A = angle C
We know that the sum of opposite angles of a parallelogram is 180.
So, angle A + angle C= 180
=> 2angle A =180
So, angleA=90.
Since a parallelogram with opposite sides equal and all angles 90 degree, is a rectangle.
Answered by
1
Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.
In ΔABC and ΔDCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
∴ ΔABC ≅ ΔDCB (By SSS Congruence rule)
⇒ ∠ABC = ∠DCB
It is known that the sum of the measures of angles on the same side of transversal is 180º.
∠ABC + ∠DCB = 180º (AB || CD)
⇒ ∠ABC + ∠ABC = 180º
⇒ 2∠ABC = 180º
⇒ ∠ABC = 90º
Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.
shreya204:
If you like my answer so please mark me as brainliest
Similar questions