Question
Two tangent segments PA and PB are drawn to a circle with centre O such that $\angle\text{APB}=120^{\circ}$ APB = 120°. Prove that OP = 2 AP.
✓
Answer
Given: From a point P. Out side the circle with centre O, PA and PB are tangerts drawn and $\angle\text{APB}=120^{\circ}$
OP is joined.
To prove: OP = 2AP
Const: Take mid point M of OP and join AM, join also OA and OB.
Proof: In right $\triangle\text{OAP},$
OP is joined.
To prove: OP = 2AP
Const: Take mid point M of OP and join AM, join also OA and OB.
Proof: In right $\triangle\text{OAP},$
$\angle\text{OPA}=\frac{1}{2}\angle\text{APB}=\frac{1}{2}\times120^{\circ}=60^{\circ}$
$\angle\text{AOP}=90^{\circ}-60^{\circ}=30^{\circ}$
M is mid point of hypotenuse OP of $\triangle\text{AOP}$
MO = MA = MP
$\angle\text{OAM}=\angle\text{AOM}=30^{\circ}$ and $\angle\text{PAM}=90^{\circ}-30^{\circ}=60^{\circ}$
$\triangle\text{AMP}$ is an equilateral triangle.
MA = MP = AP
But M is mid point of OP
OP = 2MP = 2AP
Hence proved.
$\angle\text{AOP}=90^{\circ}-60^{\circ}=30^{\circ}$
M is mid point of hypotenuse OP of $\triangle\text{AOP}$
MO = MA = MP
$\angle\text{OAM}=\angle\text{AOM}=30^{\circ}$ and $\angle\text{PAM}=90^{\circ}-30^{\circ}=60^{\circ}$
$\triangle\text{AMP}$ is an equilateral triangle.
MA = MP = AP
But M is mid point of OP
OP = 2MP = 2AP
Hence proved.
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