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A point P lies outside the circle with c...

A point P lies outside the circle with centre O. two lines PAOB and PDC are drawn on the circle from P. if PD =OD, then prove that arc `BC=3xxArcAD`.

Text Solution

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In `Delta DOP`,
`DO=DP`
`rArrangle DPO=angleDOP`
Let `angleDPO= angleDOP=x`
Now in `DeltaODP`,
`angleODC` is the exterior angle
`therefore angle ODC=angleDOP+angle DPO`
`x+x`
`2x`
In `DeltaOCD`,
`OC=OD`
`rArrangleODC=angleOCD =2x`
`thereforeangle COD=180^@-angle ODC -angle OCD`
`180^@-2x-2x`
`180^@-4x`
Now, `angleBOC+angleCOD +angle DOP=180^@` (L.P.A)
`rArrangle BOC+180^@-4x+x=180^@`
`rArrangle BOC=3x`
`therefore (angleBOC)/(angleDOP) =(3x)/(x)`
`rArr(angleBOC)/(angleDOA)=3`
`rArr(ArcBC)/(Arc AD)=3` (length of arcs are proportional to their corresponding angles )
`rArr Arc BC=3xxArcAD`
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