Fig. 6.15
In Fig. 6.16, if x+y=w+z, then prove that AOB
is a line.
B
r.r. I
Fig. 6.16​

2 thoughts on “Fig. 6.15<br />In Fig. 6.16, if x+y=w+z, then prove that AOB<br />is a line.<br />B<br />r.r. I<br />Fig. 6.16​”

1. Step-by-step explanation:

   s Given, x+y=w+z

   To Prove: AOB is a line or x+y=180

   ∘

   (linear pair.)

   According to the question,

   x+y+w+z=360

   ∘

   ∣ Angles around a point.

   (x+y)+(w+z)=360

   ∘

   (x+y)+(x+y)=360

   ∘

   ∣ Given x+y=w+z

   2(x+y)=360

   ∘

   (x+y)=180

   ∘

   Hence, x+y makes a linear pair.

   Therefore, AOB is a straight line.

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2. [tex]\huge{\textbf{\textsf{{\color{navy}{An}{\purple{swe}}{\pink{r}}{\color{pink}:)}}}}}[/tex]

   Sum of all angles in a circle always 360°

   Hence

   ∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°

   => x + y + w + z = 360°

   => x + y + x + y = 360°

   Given that x + y = w + z

   Plug the value we get

   => 2w + 2z = 360°

   => 2(w + z) = 360°

   w + z = 180° (linear pair)

   or ∠BOD + ∠AOD = 180°

   If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line

   Hence AOB is a line.

   Reply