.S4. In Fig. 6.16, if x+y=w+z, then prove that AOBFiesis a line.

1.	.S4. In Fig. 6.16, if x+y=w+z, then prove that AOBFiesis a line.
Answer» ong>Answer ⤵️ Answer: Given: (x + y) = (w + z) To prove: ∠AOB Proof: As, Sum of all ANGLES in a CIRCLE is always 360° So, according to the question, we GET: ∠AOC + ∠BOC + ∠DOB + ∠AOD = 360° => x + y + w + z = 360° => x + y + x + y = 360° (given = (x+y = w+z) so we can use x+y in the place of w+z because they are equal to each other as given) Adding the value we get, => 2x + 2y = 360° = 360°=> 2(x + y) = 360° => x + y = 360°/2 => x + y = 180° (linear pair) or ∠AOC + ∠BOC = 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.

Answer»

ong>Answer ⤵️

Answer: Given: (x + y) = (w + z)

To prove: ∠AOB

Proof: As, Sum of all ANGLES in a CIRCLE is always 360°

So, according to the question, we GET:

∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°

=> x + y + w + z = 360°

=> x + y + x + y = 360°

(given = (x+y = w+z) so we can use x+y in the place of w+z because they are equal to each other as given)

Adding the value we get,

=> 2x + 2y = 360°

= 360°=> 2(x + y) = 360°

=> x + y = 360°/2

=> x + y = 180° (linear pair)

or ∠AOC + ∠BOC = 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.

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