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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, |
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Answer» 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. |
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