{"id":1587,"date":"2026-01-03T03:33:51","date_gmt":"2026-01-02T22:33:51","guid":{"rendered":"https:\/\/rileymath.com\/?p=1587"},"modified":"2026-01-03T03:39:31","modified_gmt":"2026-01-02T22:39:31","slug":"euclids-elements-book-1-p13","status":"publish","type":"post","link":"https:\/\/rileymath.com\/index.php\/2026\/01\/03\/euclids-elements-book-1-p13\/","title":{"rendered":"Euclids Elements Book 1 P13"},"content":{"rendered":"\t\t
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\u00baIts been awhile since I have posted.  I have been busy wrapping up things with my teaching career and beginning retirement.  I am now semi-retired.  Even though I am no longer a full time teacher, I have dedicated most of my time to my former high school as a substitute, a math tutor, a music director and a subtle math coach.  By math coach, I mean that I offer help with planning to try out activities and assist core math teachers with planning.<\/p>


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So now I am getting back into an old plan to walk through Euclid\u2019s Elements one proposition at a time also watch for activities and lesson plans for second semester Algebra I.  I will reenter the ancient text in the first book at proposition 13.  Euclid stated this proposition as \u201c If a straight line set up on a straight line makes angles, it will make either two right angles or angles equal to two right angles\u201d.  Of course, he is referring to a linear pair equaling 180 degrees.  And in his typical Euclidian way, he refuses to say 180 degrees, but rather says \u201ctwo right angles\u201d.<\/p>


<\/p>\n

OK lets look at some drawings:<\/p>\n

In figure 1 Euclid stated the objective as follows:  \u201c I say that the angles CBA, ABD are either two right angles or add up to two right angles.  In Modern Geometric Notation, we would say:<\/p>\n

m\u2220ABC + m\u2220DBA = 180\u00ba<\/p>\n

\n<\/p>

\"\"<\/p>\n

\n<\/p>

This may seem obvious and not worthy of a proof, but in order to use it in further proofs, we must do a rigorous objective proof.  So here is how Euclid did it.<\/p>\n

\n<\/p>

\n<\/p>

If angle CBA is equal to ABD, they are two right angles (180\u00b0<\/span>).  But, if not, then we need figure 2 where line segment BE is drawn perpendicular to CD.<\/p>\n


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\"\"
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<\/p>\n

From Figure 2:<\/o:p><\/p>

 <\/o:p><\/p>

(1)   <\/span><\/span><\/span>m\u2220<\/span>CBE + m<\/span>\u2220<\/span>EBD = 180<\/span>\u00b0<\/span><\/p>

<\/o:p><\/p>

 <\/o:p><\/p>

and<\/o:p><\/p>

 <\/o:p><\/p>

(2)   <\/span><\/span><\/span>m\u2220<\/span>CBA + m<\/span>\u2220<\/span>ABE = m<\/span>\u2220<\/span>CBE<\/span><\/p>

<\/o:p><\/p>

 <\/o:p><\/p>

Therefore, By Subtitution:<\/o:p><\/p>

 <\/o:p><\/p>

(3)   <\/span><\/span><\/span>m\u2220<\/span>CBA + m<\/span>\u2220<\/span>ABE + m<\/span>\u2220<\/span>EBD = m<\/span>\u2220<\/span>CBE + m<\/span>\u2220<\/span>EBD = 180<\/span>\u00b0<\/span><\/p>

<\/o:p><\/p>

 <\/o:p><\/p>

We can also say:<\/o:p><\/p>

 <\/o:p><\/p>

(4)   <\/span><\/span><\/span>m\u2220<\/span>DBE + m<\/span>\u2220<\/span>EBA = m<\/span>\u2220<\/span>DBA<\/span><\/p>

<\/o:p><\/p>

 <\/o:p><\/p>

Now add m\u00d0<\/span><\/span>CBA to each side of (4)<\/o:p><\/p>

 <\/o:p><\/p>

(5)   <\/span><\/span><\/span>m\u2220<\/span>CBA+ m<\/span>\u2220<\/span>ABE + m<\/span>\u2220<\/span>EBD = m<\/span>\u2220<\/span>DBA + m<\/span>\u2220<\/span>CBA<\/span><\/p>

<\/o:p><\/p>

 <\/o:p><\/p>

Combining (3) and (5)<\/o:p><\/p>

 <\/o:p><\/p>

(6)   <\/span><\/span><\/span>m\u2220<\/span>CBE + m\u2220<\/span>EBD = m\u2220<\/span>DBA + m\u2220<\/span>CBA = 180\u00b0<\/span><\/span><\/o:p><\/p>

 <\/o:p><\/p>

Rewriting (6) <\/o:p><\/p>

 <\/o:p><\/p>

(7)   <\/span><\/span><\/span>m\u2220<\/span>DBA + m\u2220<\/span>ABC = 180\u00b0<\/span><\/span><\/o:p><\/p>

 <\/o:p><\/p>

QED<\/o:p><\/p>