$\textbf{Question:}$ Why can we drop a perpendicular $h$ between the two points along the base of the triangle below where $\alpha, \beta\leq 90^\circ$?

enter image description here

The reason I'm wondering this is because often times when dealing with isosceles triangles, I see $h$ is often constructed by connecting $M$ to the top vertex by defining $M$ to be the midpoint which implies the angle along the base is indeed ninety degrees by using the perpendicular bisector theorem. So, do I need to define where $M$ is first for this example to show construction of $h$ will be indeed $90^\circ$ or no? That's where I am getting confused..

Here is a picture.

enter image description here

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    $\begingroup$ So the question is 'why can we draw an altitude' or 'how do you draw an altitude precisely'? I don't really understand it... $\endgroup$ – Dr. Mathva Feb 9 '19 at 18:00
  • $\begingroup$ Why can we draw an altitude. Why is it feasible? Is it due to betweenness? I'm not concerned about how to draw it with a compass and a straight edge. $\endgroup$ – W. G. Feb 9 '19 at 18:02
  • $\begingroup$ I DON'T want the compass. $\endgroup$ – W. G. Feb 9 '19 at 18:05
  • $\begingroup$ What do you think geometry is about? Why do you reject the compass and straightedge? $\endgroup$ – David K Feb 9 '19 at 18:07
  • $\begingroup$ I'm not concerned about the construction of it by an example. I want the proof. $\endgroup$ – W. G. Feb 9 '19 at 18:11

Let we say about $\Delta ABC$ and let $M$ placed on the line $AB$ such that $A$ placed between $M$ and $B$.

Thus, $\measuredangle CAM<90^{\circ},$ which gives $\alpha>90^{\circ},$ which is a contradiction.

By the same way we can get a contradiction if $B$ placed between $M$ and $A$.

Id est, $M$ placed on the side $AB$.

  • $\begingroup$ Rather, we always have $\measuredangle CAM<90^{\circ}$, but if $A$ were between $M$ and $B$, $\angle CAB$ would be the supplement of $\angle CAM$ rather than the same angle. $\endgroup$ – Misha Lavrov Feb 9 '19 at 18:19
  • $\begingroup$ @Misha Lavrov If $M\in AB$ it's not always so! For example: $M\equiv A$. We have no an angle! $\endgroup$ – Michael Rozenberg Feb 9 '19 at 18:23
  • $\begingroup$ So, I made a picture above. Why do we know $\measuredangle CAM<90^\circ$? $\endgroup$ – W. G. Feb 9 '19 at 18:35
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    $\begingroup$ @W. G Because $\Delta CAM$ is a right-angled triangle. $\endgroup$ – Michael Rozenberg Feb 9 '19 at 18:36
  • $\begingroup$ Ohhhhhhh!!!! I get it thank you! $\endgroup$ – W. G. Feb 9 '19 at 18:38

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