Sedrakyan's

Trapezoid law, Trapezoid theorem and their applications

Hayk Sedrakyan, Arno Mikaelyan, Advik K.

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Abstract

In this paper, we establish some novel results. In particular: a generalization of parallelogram law (we call it Trapezoid law), and a stronger result (diagonal formulas for expressing each diagonal of a trapezoid separately using its side lengths, we call it a Trapezoid theorem). We also investigate their applications.

Introduction

Consider the following trapezoid with side lengths aa, bb, cc, dd and diagonals ee, ff.

Trapezoid with sides a, b, c, d and diagonals e, f

Trapezoid law (generalization of parallelogram's law).

We have

a2+c2=e2+f2−2bd.a^{2} + c^{2} = e^{2} + f^{2} - 2bd.

Trapezoid theorem (stronger result).

We have

e2=bd+c2b−a2db−d,e^{2} = bd + \frac{c^{2}b - a^{2}d}{b-d},
f2=bd+a2b−c2db−d.f^{2} = bd + \frac{a^{2}b - c^{2}d}{b-d}.

Remark 1. Note that Trapezoid law is the generalization of Parallelogram law. Recall that the Parallelogram law states that

2a2+2b2=e2+f2.2a^{2} + 2b^{2} = e^{2} + f^{2}.
Parallelogram with sides and diagonals

The trapezoid below (see the figure) becomes a parallelogram, if b=db=d. In that case, we get a=ca=c.

Trapezoid with sides a, b, c, d and diagonals e, f

Trapezoid law states that

a2+c2=e2+f2−2bd.a^{2} + c^{2} = e^{2} + f^{2} - 2bd.

If we take b=db=d and a=ca = c, then we get

2a2=e2+f2−2b2.2a^{2} = e^{2} + f^{2} - 2b^{2}.

Therefore, we get the Parallelogram law.

2a2+2b2=e2+f2.2a^{2} + 2b^{2} = e^{2} + f^{2}.

Remark 2. In the literature, there are 2 different definitions of a trapezoid. The first definition states that trapezoid is a quadrilateral that has exactly one pair of parallel sides, while the second definition states that trapezoid is a quadrilateral that has at least one pair of parallel sides. Trapezoid law shows that a parallelogram can actually be considered as a particular case of a trapezoid, therefore it looks like that the second definition is more appropriate as a definition of a trapezoid.

Main results and proofs

Consider the following trapezoid with side lengths aa, bb, cc, dd and diagonals ee, ff.

Trapezoid with sides a, b, c, d and diagonals e, f

Trapezoid law (generalization of parallelogram's law).

We have

a2+c2=e2+f2−2bd.a^{2} + c^{2} = e^{2} + f^{2} - 2bd.

Proof. Let us start by drawing AF parallel to BD and DE parallel to AC.

Auxiliary construction for the proof of the Trapezoid law

Using Stewart's theorem for triangles AFC and DBE, we get

a2=e2⋅d+f2⋅bd+b−d⋅b,a^{2} = \frac{e^{2}\cdot d + f^{2} \cdot b}{d+b} - d \cdot b,
c2=e2⋅b+f2⋅dd+b−d⋅b.c^{2} = \frac{e^{2}\cdot b + f^{2}\cdot d}{d+b} - d\cdot b.

Summing up the last two equations, we get

a2+c2=e2+f2−2bd.a^{2} + c^{2} = e^{2} + f^{2} - 2bd.

This ends the proof.

Trapezoid theorem (stronger result).

We have

e2=bd+c2b−a2db−d,e^{2} = bd + \frac{c^{2}b - a^{2}d}{b-d},
f2=bd+a2b−c2db−d.f^{2} = bd + \frac{a^{2}b - c^{2}d}{b-d}.

Proof. From the equations of Stewart's theorem on previous page for triangles AFCAFC and DBEDBE, we get

f2=(d+b)(db+c2)−e2bd=(d+b)(db+a2)−e2db.f^{2} = \frac{(d+b)(d b+c^{2}) - e^{2}b}{d} = \frac{(d+b)(db+a^{2}) - e^{2}d}{b}.

Thus, it follows that

(d+b)(db+c2)−e2bd−(d+b)(db+a2)−e2db=0.\frac{(d+b)(d b+c^{2}) - e^{2}b}{d} - \frac{(d+b)(db+a^{2}) - e^{2}d}{b} = 0.

Therefore

(d+b)(db+c2)d−(d+b)(db+a2)b=e2bd−e2db.\frac{(d+b)(d b+c^{2})}{d} - \frac{(d+b)(db+a^{2})}{b} = \frac{e^{2}b}{d} - \frac{e^{2}d}{b}.

We get

(d+b)(db+c2)d−(d+b)(db+a2)b=e2(b2−d2)db.\frac{(d+b)(d b+c^{2})}{d} - \frac{(d+b)(db+a^{2})}{b} = \frac{e^{2}(b^{2} - d^{2})}{db}.

Hence

b(d+b)(db+c2)−d(d+b)(db+a2)=e2(b−d)(b+d).b(d+b)(db+c^{2}) - d(d+b)(db+a^{2}) = e^{2}(b-d)(b+d).

Let us cancel d+bd+b, then

b(db+c2)−d(db+a2)=e2(b−d).b(db+c^{2}) - d(db + a^{2}) = e^{2}(b-d).

Therefore

e2=b(db+c2)−d(db+a2)b−d.e^{2} = \frac{b(db+c^{2}) - d(db+a^{2})}{b-d}.

Thus

e2=bd(b−d)−(a2d−c2b)b−d=bd+c2b−a2db−d.e^{2} = \frac{bd(b-d) - (a^{2}d - c^{2}b)}{b-d} = bd + \frac{c^{2}b - a^{2}d}{b-d}.

So, we get

e2=bd+c2b−a2db−d.e^{2} = bd + \frac{c^{2}b - a^{2}d}{b-d}.

Now, we do the same with for ff

e2=(d+b)(db+a2)−f2bd=(d+b)(db+c2)−f2db.e^{2} = \frac{(d+b)(db+a^{2})-f^{2}b}{d} = \frac{(d+b)(db+c^{2}) - f^{2}d}{b}.

We obtain

(d+b)(db+a2)−f2bd−(d+b)(db+c2)−f2db=0.\frac{(d+b)(db+a^{2})-f^{2}b}{d} - \frac{(d+b)(db+c^{2}) - f^{2}d}{b} = 0.

Hence

(d+b)(db+a2)d−(d+b)(db+c2)b=f2bd−f2db.\frac{(d+b)(db+a^{2})}{d} - \frac{(d+b)(db+c^{2})}{b} = \frac{f^{2}b}{d} - \frac{f^{2}d}{b}.

Thus

(d+b)(db+a2)d−(d+b)(db+c2)b=f2(b2−d2)db.\frac{(d+b)(db+a^{2})}{d} - \frac{(d+b)(db+c^{2})}{b} = \frac{f^{2}(b^{2}-d^{2})}{db}.

We get

b(d+b)(db+a2)−d(d+b)(db+c2)=f2(b−d)(b+d).b(d+b)(db+a^{2}) - d(d+b)(db+c^{2}) = f^{2}(b-d)(b+d).

Let us cancel d+bd+b

b(db+a2)−d(db+c2)=f2(b−d).b(db+a^{2}) - d(db+c^{2}) = f^{2}(b-d).

Therefore

f2=b(db+a2)−d(db+c2)b−d.f^{2} = \frac{b(db+a^{2}) - d(db+c^{2})}{b-d}.

Thus, it follows that

f2=bd(b−d)+a2b−c2db−d=bd+a2b−c2db−d.f^{2} = \frac{bd(b-d)+a^{2}b-c^{2}d}{b-d} = bd + \frac{a^{2}b - c^{2}d}{b-d}.

So, we get

f2=bd+a2b−c2db−d.f^{2} = bd + \frac{a^{2}b - c^{2}d}{b-d}.

This ends the proof.

Applications

Application 1 (Sedrakyan, The Stair-Step Approach in Mathematics). Consider a trapezoid with bases equal to 3 and 7. Given that the sum of the squares of its diagonals is equal to 100. Find the sum of the squares of its legs.

Solution. Let us consider the following trapezoid (see figure below).

Trapezoid for Application 1

By trapezoid law, we have that

AB2+CD2=e2+f2−2⋅BC⋅AD.AB^{2} + CD^{2} = e^{2} + f^{2} - 2\cdot BC\cdot AD.

We get

AB2+CD2=100−42=58.AB^{2} + CD^{2} = 100 - 42 = 58.

Application 2. Let ABCD be a trapezoid, where AB = 13, BC = 7, CD = 15, AD = 21 (see the figure). What is the length of diagonals AC and BD?

Trapezoid ABCD for Application 2

Solution. To find diagonal ACAC we use the Trapezoid theorem

AC2=7⋅21+152⋅7−132⋅217−21=288.AC^{2} = 7\cdot21 + \frac{15^{2}\cdot7 - 13^{2}\cdot21}{7-21} = 288.

So, we get that

AC2=288,  so  AC=122.AC^{2} = 288 ,\ \ \text{so} \ \ AC = 12\sqrt{2}.

To find diagonal BDBD we use the Trapezoid theorem

BD2=7⋅21+132⋅7−152⋅217−21=400.BD^{2} = 7\cdot21 + \frac{13^{2}\cdot7 - 15^{2}\cdot21}{7-21}= 400.

Thus, it follows that

BD2=400,  so  BD=20.BD^{2} = 400, \ \ \text{so} \ \ BD = 20.

References

  1. Sedrakyan H., Sedrakyan N., 12 Independent Practice Tests for AMC 8 Preparation: An Unofficial Study Guide, USA (2026).
  2. Sedrakyan H., Sedrakyan N., 12 Independent Practice Tests for AMC 10 Preparation: An Unofficial Study Guide, USA (2026).
  3. Sedrakyan H., Sedrakyan N., 12 Independent Practice Tests for AMC 12 Preparation: An Unofficial Study Guide, USA (2026).
  4. Sedrakyan H., Sedrakyan N., Mathematics Competitions Must-Know Geometry: Practice and theory inspired by the style of AMC and AIME geometry: An Unofficial Study Guide, USA (2023).
  5. Sedrakyan H., Sedrakyan N., 12 Independent Practice Tests for AIME Preparation: An Unofficial Study Guide, USA (2022).
  6. Sedrakyan H., Sedrakyan N., Competition math for middle school: must-knows and beyond, USA (2023).
  7. Sedrakyan H., Sedrakyan N., How to prepare for math Olympiads, USA (2019).

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