# Help show angles at corner of a pyramid add up to more than $\pi$. (Picture included)

How can I prove that $\delta_i + \gamma_{i + 1} + \beta_{i + 1} \ge \pi$? Intuitively it seems clear because if you flatten the edge of the pyramid, you are going to have to make either $\delta_i$ or $\gamma_{i + 1}$ smaller. But my brain has not had any luck supplying a proof, although I know it must be more or less trivial. Can anyone help me out?

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Hint:

Project (important: use orthogonal projection) the line that creates angles $\delta_i$ and $\gamma_{i+1}$ onto a plane that has angle $\beta_{i+1}$ and call the new angles $\delta_i'$ and $\gamma_{i+1}'$. Now observe that $\delta_i \geq \delta_i'$, $\gamma_{i+1} \geq \gamma_{i+1}'$ and $\delta_i' + \gamma_{i+1}'+\beta_{i+1} = \pi$.

I hope this helps ;-)

Edit:

Yes, if you use the orthogonal projection then the angles will be smaller. Consider the following picture:

where BC is the orthogonal projection of BD and both the red $\triangle ABD$ and blue $\triangle ABC$ triangles are right-angled (you can pick $A$ in a suitable way). Surely $|AD| \geq |AC|$ (Pythagorean theorem), but $AB$ is common to both, so $|\angle ABD| \geq |\angle ABC|$.

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This is what I have been trying to do, but showing the projected angles are smaller escapes me. Is it true that a projected angle as always smaller? I was thinking that a pyramid with $\delta_i \ge \pi/2$ might have the projected angle be larger, but still the sum $\delta_i' + \gamma_{i + 1}' \le \delta_i + \gamma_{i + 1}$. – zrbecker Jul 12 '13 at 8:49
Actually I just found out I had an inequality flipped in some computation I had done. Turning it around cleared just about everything up. – zrbecker Jul 12 '13 at 9:01
@user19536 See the edit. – dtldarek Jul 12 '13 at 9:06
Thanks I just read. I am reworking some of what I did, and for the case where the angle is less than $\pi/2$ I essentially did something similar with cosine instead of the sine side of things. However, when $D$ does not occur direction over the base of the pyramid, the angle $\angle ABD$ can be greater than $\pi/2$, in this case is it not true that the projected angle is larger? – zrbecker Jul 12 '13 at 9:14
I'm not sure what do you mean. In this consideration there is no base of a pyramid, just a plane and two lines. $C$ might be out of the base of your pyramid, but it is not relevant here. If you worry that your angles will "flip" to the other side, then you can handle this as a special case, it should be even easier than the first one. – dtldarek Jul 12 '13 at 9:24