geometry
Prove that the altitudes of an acute triangle intersect inside the triangle.
I can pretty easily see that this is true by a pythagorean theorem argument. Given any two sides, the smaller length to the side will be closer on the base so thus the orthocenter must lie inside the triangle. What would be a more mathematical way of proving this?
Note that in the case of an acute triangle you can align any side of the triangle with a side of the rectangle, and obtain the same minimal rectangle area, namely twice the area of the triangle.
I'm referring to the following figure:
We turn a circumscribing rectangle clockwise around the triangle so that when $\phi=0$ a side of the rectangle coincides with the side $AB$ of the triangle. The area $F$ of the rectangle then computes to $$F=bc\cos\phi\cos\theta={bc\over2}\bigl(\cos(\phi+\theta)+\cos(\phi-\theta)\bigr)={bc\over2}\bigl(\sin\alpha+\cos(\phi-\theta)\bigr)\ .$$ This is obviously smallest if $|\phi-\theta|$ is largest, namely $={\displaystyle{\pi\over2}}-\alpha$. In this way we obtain $$F_{\min}=bc\sin\alpha=2{\rm area}(\triangle)\ ,$$ as claimed.
Let the isoceles triangle be denoted by $\Delta ABC$, with the vertex at point $B$. Let the square be denoted by $\square DEFG$, with the vertices $D,E$ lying on the sides of the triangle and the vertices $F,G$ lying on the base. Let $r$ denote the length of the side of the square.
Note that $\Delta ABC$ is similar to $\Delta DBE$, so since the height of $\Delta ABC$ is $12$ and the height of $\Delta DBE$ is $12 - r$, we have $$\frac{r}{10} = \frac{12-r}{12}.$$ Solving for $r$ gives $r = 12\cdot 10/22 = 60/11$.
Best Answer
By contradiction:
$\triangle ABC$ is acute, $\overline{BD}$ is the altitud of the side $\overline{AC}$, then by definition $m\angle BDC=90°$, and in $\triangle ABC$, $m\angle A+m\angle ABC+m\angle ACB=180°$, in $\triangle CBD$, $m\angle D+m\angle DCB+m\angle CBD=180°\implies m\angle DCB+m\angle CBD<180 $.
By exterior angle $m\angle DCB=m\angle A+m\angle ABC$, now how $\angle ACB$ is acute, then $m\angle ACB<m\angle BDC=90°$, Then
$$m\angle A+m\angle ABC+m\angle ACB<m\angle A+m\angle ABC+m\angle BCD\implies m\angle A+m\angle ABC+m\angle BCD>180$$ then $$m\angle A+m\angle ABC+m\angle BCD=m\angle DCB+m\angle CBD>180$$
this a contradiction