Question

prove that in a parallelogram the inspectors of any two consecutive angles intersect at right angles.​

Answer 1

Step-by-step explanation:

Let ABCD is a parallelogram

as we know

∠A+∠B=180

0

OAbisects∠DAB&OBbisects∠CBA

toprove∠AOB=90

0

nowinΔAOB

∠OAB+∠CBA+∠AOB=180

0

2

1

∠DAB+

2

1

∠CBA+∠AOB=180

0

2

1

(∠DAB+CBA)+∠AOB=180

0

2

1

×180

0

+∠AOB=180

0

90

0

+∠AOB=180

0

∠AOB=180

0

−90

0

∠AOB=90

0

Answer 2

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

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, $\angle$ADB + $\angle$BCD = 180° ( From ( i ) )

So,

= $\dfrac{1}{2}$$\angle$ADB + $\dfrac{1}{2}$$\angle$BCD = $\dfrac{1}{2}$ × 180

= $\dfrac{1}{2}$$\angle$ADB + $\dfrac{1}{2}$$\angle$BCD = 90°

That is,

= $\angle$1 + $\angle$2 = 90 . . .(ii)

Now, in $\triangle$COD

= $\angle$1 + $\angle$2 + $\angle$COD = 180° { Prop. of triangle }

= 90 + $\angle$COD = 180°

= $\angle$COD = 180° - 90

= $\angle$COD = 90°

Hence Proved!!