# 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 $a = \frac{b_1 – \sqrt{b_2^2 – 4h^2}}{2}$ Clio Corvid