# Assuming that the sum of the angles of any triangle is 180, how can I deduce Euclid's 5th postulate?

I already did the reverse, namely, if we assume Euclid's 5th postulate, then the sum of the angles of any triangle is 180 degrees. Now I need to show the converse, but I don't really know how to approach this way. Any tips on how to get started would be appreciated.

Euclid's Fifth Postulate: If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

• What was Euclid's fifth postulate again? – mvw May 1 '16 at 3:55
• Euclid's original formulation was a lot more ungainly, as I recall, but the parallel postulate is equivalent. – Brian Tung May 1 '16 at 5:15
• @Chad is the formulation of the fifth postulate that I added acceptable, or is there a particular formulation you're interested in – Stella Biderman May 1 '16 at 5:27
• @StellaBiderman I know they are equivalent, but we mostly use the one of the person who answered below, thus the edit. – Chad May 1 '16 at 20:02

## 3 Answers

There is some confusion about these questions. So I state here some facts:

1. The axioms of Euclid (minus 5) prove that parallels exist.
2. Playfair's postulate is that through a point not on a line there is ONLY ONE parallel. This is equivalent to the statement of Euclid's fifth postulate.
3. $\Delta=\pi$ (that is the sum of the angles of a triange is $\pi$ radians) DOES NOT imply the parallel postulate without the Archimedean axiom.
4. If one assumes that Archimedean axiom then $\Delta=\pi \Rightarrow$Playfair.

The proof of the last statement is as follows: Let $A$ be a point not on line $l$, drop perpendicular from $A$ to $l$ at $B$. Take the line $m$ through $A$ perpendicular to $AB$, this is parallel to $l$ as Euclid shows. If there were a second line through $A$ parallel to $l$, say making angle $\beta$ with $m$, then take a point $C$ on $l$ such that $\angle ACB <\beta$ (this is where the Archimedean axiom is used). Now the right angle triangle $ABC$ as angle sum $<\pi$ as one sees form the picture.

You can construct the parallel line. Draw a line from the given point cutting the given line. Construct a line through the given point making the alternate angles, $a$ in the diagram match. Now if the lines meet to the left, the sum of the angles in the triangle exceed $\pi$. Similarly if they meet to the right the sum exceeds $\pi$. Hence the lines do not meet and are parallel.

Assume that the sum of the angles of any triangle is $180$ degrees.

Let $L_1$, $L_2$ be two straight lines that intersect in the angle $\gamma > 0$. Now, let $\bar{L_3}$ be a line segment that intersects both $L_1$ and $L_2$. Assume that $L_1$ and $L_2$ meet on the side of $\bar{L_3}$ where the two interior angles:$\alpha, \beta$ on that same side formed by $\bar{L_3}$ intersecting both $L_1$ and $L_2$, add to form a sum that is greater than or equal to $180$ degrees. Then the three angles $\alpha$, $\beta$, $\gamma$ form a triangle whose angle sum is strictly greater than $180$ degrees.

Thus, $L_1$ and $L_2$ meet on the side where the two side-sharing interior angles formed by the intersection of $\bar{L_3}$ with both $L_1$ and $L_2$ sum to less than $180$ degrees.

• I'm not sure how this answers the question. The OP is talking about the parallel postulate: " – Stella Biderman May 1 '16 at 5:12
• The OP must have updated the question while I was typing my answer. – Rustyn May 1 '16 at 5:22
• The only edit was me adding the text of the postulate – Stella Biderman May 1 '16 at 5:22
• I am not using that definition, I am using this one: If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles. – Rustyn May 1 '16 at 5:23
• Please explain downvote – Rustyn May 3 '16 at 3:47