Theorem and Proof of angle sum property of a quadrilateral.

**Prove that the sum of all the four angles of a quadrilateral is 360°.****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°.

Solved examples of angle sum property
of a quadrilateral:

**1.** The angle of
a quadrilateral are (3x + 2)°, (x – 3), (2x + 1)°, 2(2x + 5)° respectively.
Find the value of x and the measure of each angle.

**Solution: **

Using angle sum property of quadrilateral, we get

(3x + 2)°+ (x – 3)° + (2x + 1)° + 2(2x + 5)°= 360°

⇒ 3x + 2 + x - 3 + 2x + 1 + 4x + 10 = 360°

⇒ 10x + 10 = 360

⇒ 10x = 360 – 10

⇒ 10x = 350

⇒ x = 350/10

⇒ x = 35

Therefore, (3x + 2) = 3 × 35 + 2 = 105 + 2 = 107°

(x – 3) = 35 – 3 = 32°

(2x + 1) = 2 × 35 + 1 = 70 + 1 = 71°

2(2x + 5) = 2(2 × 35 + 5) = 2(70 + 5) = 2 × 75 = 150°

Therefore, the four angles of the quadrilateral are 32°, 71° 107°, 150° respectively.

**2.** In a
quadrilateral PQRS, PQ + QR + RS + SP < 2 (PR + QS).

**Solution: **

In ∆POS, PO + OS > PS …………… (i)

In ∆SOR, SO + OR > SR …………… (ii)

In ∆QOR, QO + OR > QR …………… (iii)

In ∆POQ, PO + OQ > PQ …………… (iv)

(i) + (ii) + (iii) + (iv) (Using triangle inequality property)

PO + OS + OS + OR + OQ + OR + OP + OQ > PS + SR + QR + PQ

⇒ 2 (OP + OQ + OR + OS) > PQ + QR + CS + DP

⇒ 2 [(OP + OR) + (OQ + OS)] > PQ + QR + CS + DP

⇒ 2 (PR + QS) > PQ + QR + RS + SP

The above examples will help us to solve various types of problems based on angle sum property of a quadrilateral.

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