# Q4. These quadrilaterals were convex. What would happen if the quadrilateral is not convex? Consider quadrilateral ABCD. Split it into two triangles and find the sum of the interior angles (Fig 3.7).

Draw a line matching points B and D.Line BD divides ABCD in two triangles .$\triangle BCD$ and $\triangle ABD$.

Sum of angles of $\triangle BCD$ = $\angle DBC + \angle BDC + \angle C = 180\degree$              $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

Sum of angles of $\triangle ABD$ = $\angle ADB + \angle ABD + \angle A = 180\degree$               $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2\right )$

Here,   $\angle ADB + \angle BDC = \angle D$

$\angle DBC + \angle ABD = \angle B$

Adding equation 1 and equation 2,

$(\angle DBC + \angle BDC + \angle C ) +$ $(\angle ADB + \angle ABD + \angle A )$  $=$ $360\degree$                    { $\angle ADB + \angle BDC = \angle D$   and

i.e. $\angle A + \angle B +\angle C +\angle D$ = $360\degree$.                                                                                    $\angle DBC + \angle ABD = \angle B$ }

The sum of the interior angles in a quadilateral, $\angle A + \angle B +\angle C +\angle D$ = $360\degree$.

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