For elliptic curve: $y^2=f(x)=4x^3-g_{2}x-g_{3}$. If $P_{1}=(p(z_{1}),p'(z_{1}))$, $P_{2}=(p(z_{2}),p'(z_{2}))\in f(x)$. How to prove the line $\overline{P_{1}P_{2}}$ intersect with $f(x)$ is $P_{3}=(p(z_{1}+z_{2}),-p'(z_{1}+z_{2}))$.
I have proved it using Veda theorem, but it is too routine. Is there some more simple method?
This can also be explained in terms of Weierstrass elliptic function, which has the addition formula: $$\tag{1}\begin{vmatrix} \wp(x) & \wp'(x) & 1 \\ \wp(y) & \wp'(y) & 1 \\ \wp(z) & \wp'(z) & 1 \end{vmatrix} = 0 \qquad \text{ when } x+y+z = 0$$ This says $(\wp(x),\wp'(x))$,$(\wp(y),\wp'(y))$ and $(\wp(x+y),-\wp'(x+y))$ are collinear, note that they all lie on $y^2 = 4x^3-g_2x-g_3$.
$(1)$ can be proved as follows: the determinant, when treated as function of $z$, is an elliptic function of order $3$, hence has $3$ zeroes, their sum is $0$ modulo the lattice. Two zeroes are $x,y$, hence the third must be congruent to $-x-y$.