Given a line AB and arbitrary point M outside this line, how to prove that one can drop a perpendicular to AB, and that such perpendicular is unique?
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Let $MC$ and $MD$ be two distinct perpendiculars, i.e. $m\angle{MCA}=m\angle{MDA}=90°$. Then, according to the corresponding angles theorem, $MC$ is parallel to $MD$. But parallel lines do not have common points (point $M$ belongs to both lines). Contradiction.
Vasili
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Great, but what about the first part of the proof? To drop a perpendicular. – Siegmeyer Sep 01 '17 at 12:18
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OK, to prove that there is at least one perpendicular we can use axiom that it's always possible to draw a circle with any center and distance. Let's draw a circle with $M$ as the center and radius equal to distance between $M$ and $AB$. Then we connect $M$ with the point of intersection of $AB$ and the circle. According to tangent radius theorem it will be a perpendicular. – Vasili Sep 01 '17 at 13:24
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Let $MC$ and $MD$ be two perpendiculars.
Thus, if $C$ and $D$ are different points
we get that a sum of measures angles of $\Delta MCD$ greater than $180^{\circ}$,
which is contradiction.
Michael Rozenberg
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Invoking the theorem on the sum of angles of a triangle makes use of the fifth postulate, which is not necessary here (see Vasya's answer). – Intelligenti pauca Aug 31 '17 at 19:27
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@Aretino I think theorems about sum angles of triangle and about parallel lines are the same things, but if you say... – Michael Rozenberg Aug 31 '17 at 19:31
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The theorem stating that two lines forming equal corresponding (or alternate) angles are parallel can be proved with Euclid's exterior angle theorem, which doesn't rely on fifth postulate. This postulate is needed to prove the converse theorem (that two parallel lines form equal alternate angles). – Intelligenti pauca Aug 31 '17 at 19:44