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Asked by Deepthi Last Modified
Answer 
Tasneem
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.
read less Deleted User
Setting the foundation right through fundamentals.
Let ABCD be a parallelogram. Let P be the point where diagonal AC and BD.
Let AC = BD. Then from property of parallelogram (diagonals bisect each other), AP = PC = PD = BP.
Consider traingle APD. AP = DP, hence it is a isoceles triangle. Thus Angle DAP and PDA are are equal.
Similarly Angle PCB and PBC are equal.
Angle APD and Angle CPB are equal (Opposite angles of intersecting lines)
Hence, Angles DAP, ADP, PCB, PBC are equal.
Similarly Angles PDC, PCD, PAB, PBA are equal.
But Angle PAB = Angle PCD. Interior alternate anlges of line interecting parallel lines.
Thus all the angles are equal and are right angle.
Rectangle is a parallelogram with all angles are right angles.
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