Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.
∠A +∠B+ ∠C + ∠D = ∠1+ ∠6+ ∠5+ ∠4+ ∠3+ ∠2
= ( ∠1+ ∠2+ ∠3) +(∠4+ ∠5+ ∠6)
= [By Angle sum property of triangle] 180°+180°
= 360°
Hence, the sum of measures of the triangles of a convex quadrilateral is 360°.
Yes, if quadrilateral is not convex then, this property will also be applied.
Let ABCD is a non-convex quadrilateral and join BD, which also divides the quadrilateral in two triangles.
Using angle sum property of triangle,
In ∆ABD, ∠1+ ∠2 + ∠3=180° …………(i)
In ∆ABD, ∠4+ ∠5 + ∠6=180° …………(ii)
Adding equation (i) and (ii),
∠1+ ∠2 + ∠3+∠4+ ∠5 + ∠6 = 360°
⟹ ∠1+ ∠2 + (∠3+ ∠4 ) + ∠5+ ∠6 = 360°
⟹ ∠A+ ∠B + ∠C+∠D = 360°
Hence proved.

Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/