TITLE: Altitude of a Triangle: Showing indeed perpendicular
QUESTION [0 upvotes]: $\textbf{Question:}$ Why can we drop a perpendicular $h$ between the two points along the base of the triangle below where $\alpha, \beta\leq 90^\circ$?


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.

REPLY [2 votes]: 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 $AB$.
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