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Kindly solve this,Prove that in a parallelogram the inspectors of any two consecutive angles intersect at right angles.

Answer»

Hey ! Here's your answer!

To prove: In a parallelogram the inspectors of any TWO consecutive angles intersect at right angles.

Proof:

Lets CONSIDER,

  • ABCD is a parallelogram.

We know that,

  • AB || CD and AB is a transversal . . . . ( i )

Now,

As we know that, \angleADB + \angleBCD = 180° ( From ( i ) )

So,

= \dfrac{1}{2}\angleADB + \dfrac{1}{2}\angleBCD = \dfrac{1}{2} × 180

= \dfrac{1}{2}\angleADB + \dfrac{1}{2}\angleBCD = 90°

That is,

= \angle1 + \angle2 = 90 . . .(ii)

Now, in \<klux>TRIANGLE</klux>COD

= \angle1 + \angle2 + \angleCOD = 180° { Prop. of triangle }

= 90 + \angleCOD = 180°

= \angleCOD = 180° - 90

= \angleCOD = 90°

Hence Proved!!


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