Question

In the figure, O is the centre of the circle. Prove that

<XOZ = 2(XZY + <YXZ)

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Answer 1

Given :-

• O is the centre of the given figure

To Prove :-

• $\mathsf{\angle{XOZ} = 2(\angle{XZY} + \angle{YXZ})}$

Prove :-

Here, note angle subtended by arc XY and following result.Arc XY subtends $\sf \angle{XOY}$ at Point 'O' and $\sf \angle{XZY}$ at Point 'Z' on the the remaining part of the circle.

$\sf \green {\therefore \angle{XOY} = 2\angle{XZY}}$..eq [1]

Again, note angle subtended by arc XY and following result.Arc XY subtends $\sf \angle{YOZ}$ at Point 'O' and $\sf \angle{YXZ}$ at Point 'X' on the the remaining part of the circle.

$\sf \green {\therefore \angle{YOZ} = 2\angle{YXZ}}$..[2]

Now, we have to add eq(1) and eq (2).

$\sf :\implies \angle{XOY} + \angle {YOZ}= 2\angle{XZY} + 2\angle{YXZ}$

$\sf :\implies \angle{XOY} + \angle {YOZ}= 2(\angle{XZY} + \angle{YXZ})$

$\sf :\implies \angle{XOY} + \angle {YOZ}= \angle{XOZ}$

$\sf:\implies \green {{\angle{XOZ} = 2(\angle{XZY} + \angle{YXZ})}}$

$\textsf{Hence, Proved..!}$

Answer 2

Answer:

Hey mate here is your answer.

We know that,angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.Consider the minor arc XY, that subtends ∠XPY at the centre and ∠XZY at the circle.Now, ∠XPY = 2 ∠XZY .(1)Consider the minor arc YZ, that subtends ∠YPZ at the centre and ∠YXZ at the circle.Now, ∠YPZ = 2 ∠YXZ (2)adding (1) and (2), wqe get ∠XPY + ∠YPZ = 2∠XZY + 2 ∠YXZ⇒∠XPZ = 2[∠XZY + ∠yxz )

hope it's helpful for you