Let O,P,Q,R,S (O being origin) be five points all lying on a circle in order. P,Q,R,S have respective position vector p,q,r,s
∆POQ.∆ROS + ∆POS.QOR =? A).∆PRQ ∆PRS B).∆POR ∆QOS C).([]OPQR)([]QQRS) D).([]PQRS)^2 {[] denotes AREA of quadrilateral AND ∆denotes area of triangle}
I m just not getting this .Can we prove this part? (p*q)•(r*s)+(p*s)•(q*r)=(p*r)•(q*s)
Assume that \(P_i = \left(r\cos \theta_i, r \sin \theta_i \right)\) and then check.
Really! That way.
Yeah. That's a vector by the way.
Yep,that's a vector and do you intend to say that i should solve for each one of them individually (like taking dot product of each of them).
Unfortunately...
I think that it is really very long way to go with. Didn't try that(obviously). I thought some STP would work out here in better way.
I haven't studied vector geometry or stuff. Here's another way: PQRS is a cyclic quadrilateral. What do we know about cyclic quadrilaterals? :D
Not much ..(the geometry part is really very boring to me) Just this,the opposite angles are supplementary.
Yeah. I don't know... check if that works, maybe?
@ganeshie8
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