How to prove that any segment whose ends lie on the two bases of a trapezoid is bisected by the midsegment of the trapezoid

Looking at the image I have attached (with MN being the midsegment of the trapezoid), For any PQ with P and Q lying on the bases of the trapezoid there exists ON with O being the point of intersection between PQ and MN. ON must be parallel to PC because ON is part of MN and PC is a part of DC, and MN and DC are parallel. And N must be the midpoint of CB because MN is the midsegment of the trapezoid, so MN must bisect CB. And since ON is parallel to PC and bisects CB, that means it is a midsegment of QBCP and must bisect PQ.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

2/3

step-by-step explanation: