sdfsdfs
didi you get anything yet?
ok one sec
yeah you have the right looking idea, just have to make it the proof
1) Quadrilateral OPQR is inscribed in circle N ----- Given 2)...
that is the last step, what you need to prove
It does not tell you that that is a parallelogram
Start with stating the given things , then only say what can be concluded from previous steps in the proof... 1) Quadrilateral OPQR is inscribed in circle N ----- Given 2) Angle O = 1/2 arc PQR and Angle Q = 1/2 arc ROP ----- inscribed angles theorem 3)...
Now you can say angle O plus angle Q = the sum of both those Arcs, i think it is called substitution of equality or somehting like that 1) Quadrilateral OPQR is inscribed in circle N ----- Given 2) Angle O = 1/2 arc PQR and Angle Q = 1/2 arc ROP ----- inscribed angles theorem 3) Angle O + Angle Q = 1/2 * (arc PQR + arc ROP) ------ substitution
And the sum of those two arcs is the full circle, 360 degrees, so you can sub that in 1) Quadrilateral OPQR is inscribed in circle N ----- Given 2) Angle O = 1/2 arc PQR and Angle Q = 1/2 arc ROP ----- inscribed angles theorem 3) Angle O + Angle Q = 1/2 * (arc PQR + arc ROP) ------ substitution 4) Angle O + Angle Q = 1/2 *(360) = 180 ------- circle 5)Angle O and angle Q are supplimentary --- def of supp angles
1) Quadrilateral OPQR is inscribed in circle N ----- Given 2) Angle O = 1/2 arc PQR and Angle Q = 1/2 arc ROP ----- inscribed angles theorem 3) Angle O + Angle Q = 1/2 * (arc PQR + arc ROP) ------ ADDITION 4) Angle O + Angle Q = 1/2 *(360) = 180 ------- substitution circle def 5)Angle O and angle Q are supplimentary --- def of supp angles
that is my version, there are different ways
similar way
yeah it works, you have another poof or anything?
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