Question

Prove that the sum of four angles of a quadrilateral ABCD is ${360}^0$.

Answer 1

Consider a quad quadrilateral ABCD.

Join QS.

To\quad prove:$\angle A+\angle B+\angle C+\angle D=360º$

Proof:Consider triangle ABD,

we have,$\angle A+\angle ABD+\angle ADB=180º⟹\left(1\right)$[Using Angle sum property of Triangle]

Similarly,triangle BCD,we have,$\angle DBC+\angle C+\angle BDC=180º⟹\left(2\right)$[Using Angle sum property of Triangle]

On adding (1) and (2),we get$\angle A+\angle ABD+\angle ADB+\angle DBC+\angle C+\angle BDC=180º+180º$

$\angle A+\angle B+\angle C+\angle D=360º$