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 PCDz = 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.

Answers (1)

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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