Question

PA and PB are tangents to a circle with centre O from point P. OP is equal to the diameter of the circle. Prove that ABP is equilateral triangle,

Answer 1

Diameter=2r
OP=2r
OC+CP=2r
CP=2r-r=r
OC=CP
C is MP of OP
`/_OAP` is right agled triangle
OP is hyperbola
OA=AC=OC=r
`/_AOC=60^0`
`/_AOP+/_APO=90^0`
`/_APO=90-60=30^0`
`/_APB=60^0`
`/_PAB=/_PBA`
`/_PAB+/_PBA+/_APB=180^0`
`2/_PAB=120^0`
`/_PAB=60^0=/_PBA`
`/_PAB` is an equilateral triangle.