Let $\triangle ABC$ be an isosceles triangle with base $a$ and altitude to the base $b.$ I am trying to find the sides of the rectangle inscribed in $\triangle ABC$ if its diagonals are parallel to the triangle legs.
Does an inscribed rectangle exist in every isosceles triangle? How are we to construct that rectangle?
Thank you in advance!
Best Answer
Yes, any number of triangles can be constructed that way. Draw a line parallel to one leg. See where it cuts the altitude. Reflect this parallel parallel line about altitude to be parallel to the other leg. Draw the inscribed rectangle as shown including cutting points on both the slant sides of the isosceles triangle.
The slant parallel lines cannot be called diagonals in general. They help to locate intersection point.. of concurrency ( slant leg, breadth and height of rectangle. )
The cutting point can be even outside the given isosceles triangle $ABC$ ( base $a$, height $b$). That is, $h>b$ possible as well as $h<0$ is possible.
Equation of slant leg
$$ \frac{2x}{a}+\frac{y}{b}=1 $$
If you plug in $ y=h$ then the $x-$ coordinate of the corner of rectangle is:
$$ x1=(1-\frac{h}{b}) \frac{a}{2}$$
When $ h>b, x1 $ goes negative, to the left of base center.