How to prove sum of angles of quadrilateral is 360?
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Consider a quadrilateral PQRS.
Join QS.
To prove: ∠P + ∠Q + ∠R + ∠S = 360º
Proof:
Consider triangle PQS, we have,
⇒ ∠P + ∠PQS + ∠PSQ = 180º ... (1) [Using Angle sum property of Triangle]
Similarly, in triangle QRS, we have,
⇒ ∠SQR + ∠R + ∠QSR = 180º ... (2) [Using Angle sum property of Triangle]
On adding (1) and (2), we get
∠P + ∠PQS + ∠PSQ + ∠SQR + ∠R + ∠QSR = 180º + 180º
⇒ ∠P + ∠PQS + ∠SQR + ∠R + ∠QSR + ∠PSQ = 360º
⇒ ∠P + ∠Q + ∠R + ∠S = 360º [Hence proved]
Join QS.
To prove: ∠P + ∠Q + ∠R + ∠S = 360º
Proof:
Consider triangle PQS, we have,
⇒ ∠P + ∠PQS + ∠PSQ = 180º ... (1) [Using Angle sum property of Triangle]
Similarly, in triangle QRS, we have,
⇒ ∠SQR + ∠R + ∠QSR = 180º ... (2) [Using Angle sum property of Triangle]
On adding (1) and (2), we get
∠P + ∠PQS + ∠PSQ + ∠SQR + ∠R + ∠QSR = 180º + 180º
⇒ ∠P + ∠PQS + ∠SQR + ∠R + ∠QSR + ∠PSQ = 360º
⇒ ∠P + ∠Q + ∠R + ∠S = 360º [Hence proved]
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Statement :
sum of the angles of quadrilateral is 360°
To Prove :
∠A + ∠B + ∠C + ∠D = 360°
Proof :
In ∆ ABC , m∠4 + m∠5+m∠6 = 180°
[ using angle a property of a triangle]
Also , in ∆ ADC , m∠1 + m∠2+m∠3= 180°
Sum of the measures of ∠A, ∠B , ∠C and ∠D of a quadrilateral
m∠4 + m∠5+ m∠6 + m∠1 + m∠2 +m∠3 = 180°+ 180°
→ ∠A + ∠B + ∠C + ∠D = 360°
Thus , sum of measure of four angles of quadrilateral is 360°.
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