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Preeti Dabral 1 month, 1 week ago

Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.

{tex}\angle \mathrm{A}+\mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}{/tex}

= {tex}\angle 1+\angle 6+\angle 5+\angle 4+\angle 3+\angle 2{/tex}

= {tex}(\angle 1+\angle 2+\angle 3)+(\angle 4+\angle 5+\angle 6){/tex}

= {tex}180^{\circ}+180^{\circ}{/tex} [By Angle sum property of triangle]

= {tex}360^{\circ}{/tex}

Hence, the sum of measures of the triangles of a convex quadrilateral is {tex}360^{\circ}{/tex}.

Yes, if quadrilateral is not convex then, this property will also be applied.

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