In the adjacent figure $- - \to O P, - - \to O Q, - - \to O R$ and $- - \to O S$ are four rays. Prove that
$∠ P O Q + ∠ Q O R + ∠ S O R + ∠ P O S = 360^{o}$ .

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Solution

In the given figure, you need to draw opposite ray to any of the rays $- - \to O P, - - \to O Q, - - \to O R$ or $- - \to O S$

Draw ray $- - \to O T$ so that $\leftarrow -- \to T O Q$ is a line. Now, ray OP stands on line $\leftarrow \to T Q$ .

Therefore, $∠ T O P + ∠ P O Q = 180^{o}$ .... (1) (Linear pair axiom)

Similarly, ray $- - \to O S$ stands on line $\leftarrow \to T Q$ .

Therefore, $∠ T O S + ∠ S O Q = 180^{o}$ .... (2) (Why?)

But $∠ S O Q = ∠ S O R + ∠ Q O R$

So, (2) becomes
$∠ T O S + ∠ S O R + ∠ Q O R = 180^{o}$ ..... (3)

Now, adding (1) and (3), you get
$∠ T O P + ∠ P O Q + ∠ T O S + ∠ S O R + ∠ Q O R = 360^{o}$ ... (4)

But $∠ T O P + ∠ T O S = ∠ P O S$

Therefore, (4) becomes
$∠ P O Q + ∠ Q O R + ∠ S O R + ∠ P O S = 360^{o}$

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