Question

o is the centre of the circle. pa and pb are tangents to the circle from point p. prove that AOBP is a cyclic quadrilateral.

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Answer 1

I'll be short.

Angle OAP = Angle OBP = 90°

[Because tangents are perpendicular to the radius at the point of contact]

Therefore, OAP+OBP = 180° which is a property of a cyclic quadrilateral but we don't know if OAPB is a cyclic quadrilateral yet.

Let's join OP.

In triangle OAP, angle OAP = 90°

This means that if we take OP as the diameter of a circle then point A lies on that circle because then angle OAP would be the angle in a semi circle arc.

BUT THIS WOULD ALSO MEAN THAT POINT B ALSO LIES ON THE SAME CIRCLE BECAUSE ANGLE OBP (= 90°) WOULD ALSO BE AN ANGLE IN A SEMI CIRCLE ARC.

Therefore, points A and B lie on a circle whose diameter is OP, i.e points A, B, O, P lie on the same circle. Hence, the quadrilateral OBPA formed by joining these points is a Cyclic Quadrilateral.

-Q.E.D

Answer 2

Step-by-step explanation:

hehe m an 11th student and it will be fun to explain my junior the same way l understood from my teacher..

See,

first draw a virtual line connecting O to P.

Now,

this divides the quadrilateral into two

equal triangles....let's prove this logic....

ln ΔOBP and ΔOAP,

<OBP = <OAP ( 90°)

OP = OP ( common)

OB = OA ( radii of a circle)

Therefore, using RHS congruency rule,

ΔOBP ≈ ΔOAP --- (1)

Now, from (1),

let <POB = <POA = x -----(2)

Also, <OPB = <OPA = y -----(3)

Then,

<OBP = <OAP = 90° ( Tangents are always

perpendicular to the

rafius of a circle)

So, <OBP + <OAP = 180° ------(4)

Lastly,

ΔOBP + ΔOAP = ◼️AOBP <OBP+<POB+<OPB+<OAP+<POA+<OPA = 360°

90° + x + y + 90° + x + y = 360° (from (2) & (3))

180° + 2x + 2y = 360°

2x + 2y = 180°

2x = <POB + <POA = <AOB

2y = <OPB + <OPA = <APB

hence, <AOB + <APB = 180° -------(5)

From (4) & (5),

AOBP is a cyclic quadrilateral.

Hence proved...