Question

Prove that sum of parallelogram in 360

Answer 1

parallelogram is a quadrilateral...
Proof: Let ABCD be a quadrilateral. Join AC. 

Clearly, ∠1 + ∠2 = ∠A ...... (i) 

And, ∠3 + ∠4 = ∠C ...... (ii) 

We know that the sum of the angles of a triangle is 180°. 

Therefore, from ∆ABC, we have

∠2 + ∠4 + ∠B = 180° (Angle sum property of triangle)

From ∆ACD, we have 

∠1 + ∠3 + ∠D = 180° (Angle sum property of triangle) 

Adding the angles on either side, we get; 

∠2 + ∠4 + ∠B + ∠1 + ∠3 + ∠D = 360° 

⇒ (∠1 + ∠2) + ∠B + (∠3 + ∠4) + ∠D = 360° 

⇒ ∠A + ∠B + ∠C + ∠D = 360° [using (i) and (ii)]. 

Hence, the sum of all the four angles of a quadrilateral is 360°.