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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!

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    $\begingroup$ Yes the rectangle exists. Just construct the diagonals through the centroid to get your rectangle. $\endgroup$
    – user10354138
    Jul 22, 2020 at 14:50
  • $\begingroup$ @user10354138, can I ask why the centroid is the intersection of the diagonals? Why are we sure? $\endgroup$
    – Katherine
    Jul 22, 2020 at 15:01
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    $\begingroup$ If you draw a diagram, you see there are three congruent triangles screaming at you --- one from the apex to the closest edge of the rectangle, the second is that edge to the intersection of diagonals, the third one is on the base. So the intersection of diagonals lie 1/3 way up the altitude, i.e., the centroid of the isoceles triangle. $\endgroup$
    – user10354138
    Jul 22, 2020 at 15:01
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    $\begingroup$ Thank you so much! My question wasn't clear though. I am asking why the intersection of diagonals lie on the altitude. $\endgroup$
    – Katherine
    Jul 22, 2020 at 15:12
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    $\begingroup$ Because the first thing to try is to have the rectangle invariant under reflection in the axis of symmetry, i.e., the altitude. So the intersection of diagonals must be on this axis. $\endgroup$
    – user10354138
    Jul 22, 2020 at 15:17

2 Answers 2

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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. )

enter image description here

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.

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enter image description here

Let $ABC$ be a an isosceles triangle as shown in the figure. $D(t,0)$ and $E$ be points on the side $AB$ and $AC$ respectively such that $DE$ is one of the diagonal of the rectangle inscribed in $\triangle ABC$.

Equation of $AC$ is given as $\displaystyle y=\left(\frac{2b}{a}\right) x+b$. It is evident by symmetry that $x$- coordinate of $E$ will be $-t$ and hence we've $y$- coordinate as $\displaystyle b-\frac{2bt}{a}$.

Given that slope of $BC$ and $DE$ are equal, we get $\displaystyle t=\frac{a}{6}$.

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