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Q3. As before consider quadrilateral ABCD (Fig 3.6). Let P be any point in its interior. Join P to vertices A, B, C and D. In the figure, consider $\Delta PAB$ . From this we see $x = 180\degree - m\angle 2 - m\angle 3$ ; similarly from $\Delta PBC$ , $y = 180\degree - m\angle 4 - m\angle 5$ , from $\Delta PCD$ , $z = 180\degree - m\angle 6 - m\angle 7$ and from $\Delta PDA$ , $w = 180\degree - m\angle 8 - m\angle 1$ . Use this to find the total measure $m\angle 1 + m\angle 2 + ... + m\angle 8$ , does it help you to arrive at the result? Remember
$\angle x + \angle y + \angle z + \angle w = 360\degree$ .

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As we know $\angle x + \angle y + \angle z + \angle w = 360\degree$

$(180\degree - m\angle 2 - m\angle 3)$ $+$ $(180\degree - m\angle 4 - m\anlge 5)+$ $(180\degree - m\angle 6 - m\angle 7) +$ $( 180\degree - m\angle 8 - m\angle 1)$ $= 360\degree$

$m\angle 1 + m\angle 2 + ... + m\angle 8$ $=$ $\left ( 180\degree+180\degree+180\degree+180\degree \right ) - 360\degree$

$m\angle 1 + m\angle 2 + ... + m\angle 8$ $= 720\degree - 360\degree$

$m\angle 1 + m\angle 2 + ... + m\angle 8$ $= 360\degree$

Hence, sum of $m\angle 1 + m\angle 2 + ... + m\angle 8$ $= 360\degree$ .

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