geometrylinear algebra
Please show me how do you find length of BH or AI (in general), I want both specific proof for this case and in a general case.
My purpose is to try to prove that the area of the square is the same as the area of the parallelogram given two parallel line L1 and L2 and the square and the parallelogram share the same base AB.
from here, $AB$ mean $\overrightarrow{AB}$.
$f$ is counterclockwise-$90^{\circ}$-rotation transformation, which is linear.
$f(MP)=f(MQ+QB+BT+TP)=f(MQ)+f(QB)+f(BT)+f(TP)$
$=QA+MQ+PT+TC=QA+MQ+RN+AR$
$=MQ+QA+AR+RN=MN$
thus, $~\square MPON$ is square.
Converting comments to an answer, as requested ...
Suppose that $a$ is the shorter (and "upper") base. Draw a line through an endpoint of the $a$-base, parallel to the opposite lateral side; this cuts the trapezoid into a parallelogram (with upper and lower bases $a$) and a triangle (with (lower) base $b-a$). The drawn line separates your "middle base", say of length $c$, into a part (of length $a$) within the parallelogram and a part (of length $c-a$) within the triangle.
If $h$ is the height of the parallelogram (and triangle), and $k$ is the height the upper sub-trapezoid (and upper sub-triangle), then, by similarity, $$\frac{k}{h}=\frac{c-a}{b-a} \tag{1}$$
Now, assuming the middle base bisects the area, so that the upper sub-trapezoid's area is half the whole, we have $$\frac{1}{2}k(a+c) = \frac12\cdot\frac12h(a+b) \quad\to\quad \frac{k}{h} = \frac12\frac{a+b}{a+c} \tag{2}$$
Setting $(1)$ and $(2)$ equal gives $$\frac{c-a}{b-a}=\frac12\frac{a+b}{a+c}\quad\to\quad 2(c^2-a^2) = b^2-a^2 \quad\to\quad 2c^2 = a^2+b^2 \tag{3}$$ which is the target relation. $\square$
Best Answer
The area of any parallelogram is base times the height and in your case it is: $AB.BH$ which is the same as the area of the square.