On circle with center O are points A, B, C and D and so that AB and CD are perpendicular diameters of circle. Point M is inisde of line segment AB. Let N be a intersection CM with circle. Tangent line at N intersect with perpendicular to AB at point M. Prove that PO=MC
>>7963845
similar triangles sharing base OM and height r.
>>7963856
but how you prove that triangle MOD have geight of r?
>>7964707
*MOP
>Given: AO=BO=CO=DO
>Line ON, because N is on the edge of the circle NO=AO
>ONP=90 because OP is tangent and NO is going through the center
>OQM=PQN because they are opposite angles of intersecting lines
>Because OMP=90 we know NOM=NPM and triangles NQP and OMQ are equal
>From that we know PN=MO
>Because we know PN=MO, NO=CO, and angles ONP=MOC we can say that OP=MC
>QED
>>7964756
>triangles NQP and OMQ are equal
How do you know that they aren't just similar?
>>7964826
not that guy, but because NP = MO, NPOM forms an isosceles trapezoid
>>7965130
Why? What tell you that it is isosceles trapezoid?
All the angles are the same and they share a side. You can use the law of sines to show they are equal
