Question

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In an circle, O is the centre of the circle. PA and PA are tangents. Show that AOBP is a cyclic quadrilateral.​

Answer 1

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A cyclic quadrilateral is one in which sum of opposite angles are supplementary.

or. Sum of opposite angles are = 180°

As point P is an external point and from P two tangents of equal length PA and PB are drawn at circle ⭕.

From the theorem of tangent, the radius drawn an angle of 90°.

Thus

→ < OBP = 90°

→ < OAP = 90°

→ < OBP + < OAP = 90°+90°

→ < OBP + < OAP = 180 °

So, <AOB + <BPA = 180°

Hence OABP is a cyclic quadrilateral.

Answer 2

Answer:

A cyclic quadrilateral is one in which sum of opposite angles are supplementary.

or. Sum of opposite angles are = 180°

As point P is an external point and from P two tangents of equal length PA and PB are drawn at circle ⭕.

From the theorem of tangent, the radius drawn an angle of 90°.

Thus

→ < OBP = 90°

→ < OAP = 90°

→ < OBP + < OAP = 90°+90°

→ < OBP + < OAP = 180 °

So, <AOB + <BPA = 180°

Hence OABP is a cyclic quadrilateral.