Isosceles Trapezoids and Right Angles

In this entry, I’m going to start with a concrete problem and develop an abstract generalization.

The starting problem: Given isosceles trapezoid $ABCD$ with an altitude of 6. Point $E$ is on $\overline{DC}$ such that $DE = 3$, $EC = 8$, and $\angle AEB$ is right. Determine $AB$.

We can solve this by placing points $G$ and $H$ on $\overline{AB}$ such that $\overline{EG}\perp\overline{AB}$ and $\overline{CH}\perp\overline{AB}$.

Since $ABCD$ is isosceles, $AF = BH = x$. We can identify three right triangles:

• $\Delta AGE$ has sides 6, $x + 3$, and $AE$
• $\Delta BGE$ has sides 6, $x + 8$, and $BE$
• $\Delta AEB$ has sides $AE$, $BE$, and $2x + 11$

Since we have three unknowns ($x, AE, BE$), it is possible for us to solve a system of three equations. These are the sides of right triangles, so we’ll use the Pythagorean Theorem:

• $AE^2 = (x + 3)^2 + 6^2 = x^2 + 6x + 9 + 36 = x^2 + 6x + 45$
• $BE^2 = (x + 8)^2 + 6^2 = x^2 + 16x + 64 + 36 = x^2 + 16x + 100$
• $AE^2 + BE^2 = (2x + 11)^2 = 4x^2 + 44x + 121$

Adding the two equations and using transitivity yields a single equation with a single unknown:

$$x^2 + 6x + 45 + x^2 + 16x + 100 = 4x^2 + 44x + 121 \Rightarrow \\ 2x^2 + 22x + 145 = 4x^2 + 44x + 121 \Rightarrow \\ 2x^2 + 22x – 24 = 0 \Rightarrow \\ x^2 + 11x – 12 = 0$$

This has the roots $x \in \{1, -12\}$. This yields $AB \in \{13, -13\}$. $AB$ has to be positive, so $AB = 13$.

Generalizing the Solution

The next level of abstraction is to determine the value of $x$ given some altitude $c$ and $E$ placed to create lengths of $a$ and $b$. Returning to our three triangles:

• $\Delta AGE$ has sides $c$, $x + a$, and $AE$
• $\Delta BGE$ has sides $c$, $x + b$, and $BE$
• $\Delta AEB$ has sides $AE$, $BE$, and $2x + a + b$

which gives us:

• $AE^2 = (x + a)^2 + c^2 = x^2 + 2ax + a^2 + c^2$
• $BE^2 = (x + b)^2 + c^2 = x^2 + 2bx + b^2 + c^2$
• $AE^2 + BE^2 = (2x + a + b)^2 = 4x^2 + 2ax + 2bx + 2ax + a^2 + ab + 2bx + ab + bx^2 \\ = 4x^2 + 4ax + 4bx + a^2 + 2ab + b^2$

So that: $$x^2 + 2ax + a^2 + c^2 + x^2 + 2bx + b^2 + c^2 = 4x^2 + 4ax + 4bx + a^2 + 2ab + b^2 \Rightarrow \\ 2x^2 + 2ax + 2bx + 2c^2 + a^2 + b^2 = 4x^2 + 4ax + 4bx + a^2 + 2ab + b^2 \Rightarrow \\ 2x^2 + 2ax + 2bx + 2ab – 2c^2 = 0 \Rightarrow \\ x^2 + (a + b)x + ab – c^2 = 0$$

The Quadratic Formula allows us to solve this for $x$: $$x = \frac{-(a + b) \pm \sqrt{(a + b)^2 – 4(ab – c^2)}}{2}$$

Simplifying the radicand yields: $$x = \frac{-(a + b) \pm \sqrt{(a – b)^2 + 4c^2}}{2}$$

Note that $x$ can be negative; that would mean that $AB < DC$, which is acceptable. However, there is the constraint that $2x > -(a + b)$ (since $AB$ has to be positive), so we can only take the larger root: $$x = \frac{-(a + b) + \sqrt{(a – b)^2 + 4c^2}}{2}$$

This will have integer solutions when the radicand ($(a – b)^2 + 4c^2)$) is a perfect square, such as all trapezoids of altitude $6$ where $|a – b| = 5$.

Valid Trapezoids

This raises the question: What are the restrictions on an isosceles trapezoid’s base and altitude that allows for a right angle to be drawn on the base as in the original image?

For trapezoid $ABCD$, drawing a diagonal $\overline{BD}$ creates angles $\angle BDA$ and $\angle BDC$.

For trapezoid $ABCD$, placing midpoint $M$ on $DC$ and drawing $\overline{AM}$ and $\overline{BM}$ creates angle $\angle BMA$.

It is always the case that $m\angle{BDA} < m\angle{BMA}$. There will be thus some point $E$ on $DC$ such that $m\angle BEA = 90^{\circ}$ if and only if $m\angle{BDA} \le 90^{\circ} \le m\angle{BMA}$.

Let’s return to an earlier equation. Rather than solving it for $x$, we’ll solve it for $a$: $$x^2 + ax + bx + ab – c^2 = 0$$

$a$ represents the distance between the trapezoid’s vertex $D$ and the collinear point $E$. For a given isosceles trapezoid, we have three basic values: the lengths of the two bases, $b_1$ and $b_2$, and the altitude $h$. $x = \frac{\Delta b}{2}$, while $b = b_1 – a$. Since $c = h$, we can rewrite the equation:  $$\frac{(\Delta b)^2}{4} + \frac{a\Delta b}{2} + \frac{(b_1 – a)\Delta b}{2} + a(b_1 – a) – h^2 = 0 \Rightarrow  \\ a^2 – ab_1 – \frac{b_1\Delta b}{2} – \frac{(\Delta b)^2}{4} + h^2 = 0 \Rightarrow a = \frac{b_1 \pm \sqrt{b_1^2 – 4\left(h^2 – \frac{b_1\Delta b}{2} – \frac{(\Delta b)^2}{4}\right)}}{2}$$

Let’s simplify the radicand. Since $\Delta b = b_2 – b_1$, $$b_1^2 – 4\left(h^2 – \frac{b_1\Delta b}{2} – \frac{(\Delta b)^2}{4}\right) = b_1^2 – 4\left(h^2 – \frac{b_1b_2 – b_1^2}{2} – \frac{b_2^2 – 2b_1b_2 + b_1^2}{4}\right) \\ = b_1^2 – 4h^2 + 2b_1b_2 – 2b_1^2 +b_2^2 – 2b_1b_2 + b_1^2 = b_2^2 – 4h^2$$

Hence, $$a = \frac{b_1 \pm \sqrt{b_2^2 – 4h^2}}{2}$$

This will be real when $b_2 \ge 2h$. Thus, one constraint on our trapezoid is that the length of its bottom base must be at least twice its altitude. If $b_2 = 2h$, then the midpoint of $b_1$ will be the vertex of a right angle with sides that contain $A$ and $B$.

The other constraint is that the lower root of $a$ must be non-negative. If $a$ is negative, then it is to the left of $D$ and hence exterior to the trapezoid. The lower root is non-negative when $b_1 \ge \sqrt{b_2^2 – 4h^2} \Rightarrow b_1^2 \ge b_2^2 – 4h^2$.

Conclusion

Putting these together, it is possible to create a right angle with vertex $E$ on base $CD$ and sides containing $A$ and $B$ for an isosceles trapezoid ABCD with these constraints: $$b_1^2 \ge b_2^2 – 4h^2 \\ b_2 \ge 2h$$

The distance of this vertex from $C$ or $D$ will be  $$a = \frac{b_1 – \sqrt{b_2^2 – 4h^2}}{2}$$