1) Over half the diameter of the circle center O M such that AB took the offer AM / supply MC. Retrieved point C on the same AM, with half the tangent at M (O) cut the line BC at K. OI vgoc up with BC. E is called the intersection of the half-OI (O).
a) CMR: KMIO inscribed quadrilateral
b) CMR: AC / / OI
c) Call F is the intersection of EM with BK. CMR: FKM weight triangle
d) CMR: KC + KB - 2KM> 0

2) For the triangle ABC has 3 sharp corners (no weight triangle in A) inscribed (O). 3 High Road AK, BE, CD intersect at H
a) CMR: BDEC inscribed quadrilateral
b) Triangle ADC with uniform triangle AEB. Deduced from AD.AB = AE.AC
c) CMR: KA is the beam angle bisector of DKE
d) Call I and J respectively, the midpoint of BC and DE
CMR: OA / / IJ