Question about Kepler's second law. I was watching a Fehnman lecture on YouTube, where he used Kepler's second law as an example of something he was explaining.
He was showing geometrically why a line joining a planet and a Sun sweeps out equal areas during equal intervals of time.
I followed his explanation for when the Sun has no pull on the planet.  Then, I'm just solving the area of obtuse angles.  
His explanation begins here and he loses me when he says that the triangular area swept out when the Sun has a force has the same height. 
I understand that both triangles have the same base, but I don't understand how to see that they're the same height. 
His explanation was, "...and do they have the same altitude?  Sure, because they're included between parallel lines and so they have the same altitude."
That doesn't explain it for me, and so can someone show me how it's done?
 A: The matter was settled in the comments, by Henning Makholm: 

in the first half of the argument we too the orbit to be the base of the triangles, and in the second hand we switch to considering the middle one of the sun-planet radii to be the base. 

and by the OP: 

I was trying to see how the base of the 3nd triangle was the same as the height of the 1st triangle. But what I really had to do was see that the base of the 3rd triangle was the same as the base of the 2nd triangle. Then I could say that the 3rd triangle had the same area as the 2nd triangle, and I already knew that the 2nd triangle had the same area as the first. 

I'll add a calculus version of this argument: placing the Sun into the origin of polar coordinates, we find that the position vector $\mathbf r$ of the planet satisfies $\ddot{\mathbf r}\times \mathbf r=0$ (force collinear to $\mathbf r$). Therefore, $$\frac{d}{dt}(\dot {\mathbf r}\times \mathbf r) = \ddot {\mathbf r}\times \mathbf r+ \dot {\mathbf r}\times \dot{\mathbf r}=0+0=0$$ 
that is, $\dot {\mathbf r}\times \mathbf r$ is constant. The magnitude of the latter vector is twice the rate at which the area is swept.
