If the diagonals of a parallelogram are equal, then show that it is a rectangle
Solution:
Given: The diagonals of a parallelogram are equal.
To show that a given parallelogram is a rectangle, we have to prove that one of its interior angles is 90° and this can be done by the concept of congruent triangles.
Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90°.
In ∆ABC and ∆DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given the diagonals are equal)
∴ ∆ABC ≅ ∆DCB (By SSS Congruence rule)
⇒ ∠ABC = ∠DCB (By CPCT) ------------- (1)
It is known that the sum of the measures of angles on the same side of transversal is 180° (co - interior angles)
∠ABC + ∠DCB = 180° (AB || CD)
⇒∠ABC + ∠ABC = 180° [From equation(1)]
⇒ 2∠ABC = 180°
⇒∠ABC = 90°
Thus, ∠DCB = 90° [From equation (1)]
Hence, ∠B = ∠D = ∠C = ∠A = 90° [Since opposite angles of a parallelogram are equal].
Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 8
Video Solution:
If the diagonals of a parallelogram are equal, then show that it is a rectangle
NCERT Maths Solutions Class 9 Chapter 8 Exercise 8.1 Question 2
Summary:
If the diagonals of a parallelogram are equal, we have proved that it is a rectangle.
☛ Related Questions:
- Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
- Show that the diagonals of a square are equal and bisect each other at right angles.
- Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
- Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that i) it bisects ∠C also, ii) ABCD is a rhombus.