Question

Prove that Sum of angles of a quadrilateral is 360° (interior) ​

Answer 1

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°.

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Answer 2

Answer:

Step-by-step explanation:

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°.