Answer:
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)
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Answer:
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)