Question

o is the center of a circle, PA and PB are tangents to a circle from point P. prove that
PAOB is a cyclic quadrilateral
PO is the bisector of angle APB
angle OAB = angle OPA

Answer 1

Answer:

Step-by-step explanation:

(i) angle PAO = angle PBO = 90

​ In quadrilateral PAOB,

​angle PAO + angle PBO + angle APB + angle AOB = 360

​90 + 90 + angle APB + angle AOB = 360

​angle APB + angle AOB = 180

​Since OPPOSITE ANGLES SUM UP TO 180, it is a CYCLIC ​QUADRILATERAL

(ii) In triangle APO and triangle BPO,

​angle PAO = angle PBO(90)

OP=OP(common)

OA=OB(radius)

​Therefore, triangle APO is congruent to triangle BPO (RHS)

​Hence, angle APO = angle BPO(cpct)

So, OP is the bisector of angle APB.

​HOPE YOU LIKE THIS ANSWER.

Answer 2

Answer: attached below