The diagonals of the inscribed quadrangle $ABCD$ intersect at point $K$. Prove that the tangent at point $K$ to the circle circumscribed around the triangle $ABK$ is parallel to $CD$.
(A Zaslavsky)
parmenides51 wrote:
The diagonals of the inscribed quadrangle $ABCD$ intersect at point $K$. Prove that the tangent at point $K$ to the circle circumscribed around the triangle $ABK$ is parallel to $CD$.
(A Zaslavsky)
Very straightforward.
We know that the radius of circle of ABK triangle through K is perpendicular to CD.
The tangent at K is also perpendicular to the radius at K.
Hence the parallel result follows
