In the figure given below, O is the centre of the circle and PA and PB are tangent segments. Show that quadrilateral AOBP is cyclic.
Answers
Answered by
113
Answer:
I am attaching the step wise solution of the given problem below. Check this solution and if you have any problem, drop a comment in the comment box.
As triangle are perpendicular to the radius so
OA ⊥AP & OB ⊥BP
∠OAP = 90°, ∠OBP = 90°
∠OAP+ ∠OBP = 180°
In quadrilateral , OAPB
∠APB +∠AOB +∠AOP +∠OBP = 360
∠APB + ∠AOB + 180 = 360
∠APB + ∠AOB = 180
I hope this helps for you.
Attachments:
![](https://hi-static.z-dn.net/files/dbd/90eb2113adb69d9425149589581878bb.jpg)
Answered by
55
Solution:
A cyclic quadrilateral is one in which sum of opposite angles are supplementary.
ie. 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,that radius drawn an angle of 90°
Thus
< OBP = 90°
< OAP = 90°
< OBP + < OAP = 90°+ 90°
< OBP + < OAP = 180°
these are the opposite angles of cyclic quadrilateral,and sum of both are 180°, thus sum of remaining two angles are also 180°
< AOB + < BPA = 180°
hence OAPB is a cyclic quadrilateral.
Hope it helps you.
A cyclic quadrilateral is one in which sum of opposite angles are supplementary.
ie. 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,that radius drawn an angle of 90°
Thus
< OBP = 90°
< OAP = 90°
< OBP + < OAP = 90°+ 90°
< OBP + < OAP = 180°
these are the opposite angles of cyclic quadrilateral,and sum of both are 180°, thus sum of remaining two angles are also 180°
< AOB + < BPA = 180°
hence OAPB is a cyclic quadrilateral.
Hope it helps you.
Attachments:
![](https://hi-static.z-dn.net/files/dbf/9cfb6861951b7fc4d0c79b46cd2da4ab.jpg)
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