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Dual Pythagorean Theorem

Dual Pythagorean Theorem

Lesson · Intermediate

Geometry: Plane Geometry

Dual Pythagorean Theorem

CB^2-CA^2 = EB^2-EA^2 = FB^2-FA^2 = DB^2-DA^2

Dual Pythagorean Theorem proof

By applying the Pythagorean Theorem to triangles $\triangle BCD$ and $\triangle ACD$

CB^2=CD^2+DB^2

CA^2=CD^2+DA^2

Therefore

CB^2-CA^2=DB^2-DA^2

Similarly

EB^2-EA^2=DB^2-DA^2

FB^2-FA^2=DB^2-DA^2

Hence

CB^2-CA^2 = EB^2-EA^2 = FB^2-FA^2 = DB^2-DA^2

Find $x$ in terms of $a,b,c$ (side lengths of $\triangle ABC$ )

Triangle ABC with altitude AD and BD equal to x

Triangle ABC with altitude h_a and base segments x and a-x

x^2+h_a^2=c^2

(a-x)^2+h_a^2=b^2

Subtracting

x^2-(a-x)^2=c^2-b^2

x^2-(a^2-2ax+x^2)=c^2-b^2

2ax-a^2=c^2-b^2

2ax=a^2+c^2-b^2

\boxed{x=\frac{a^2+c^2-b^2}{2a}}

Find $x$

Rectangle ABCD with point P and distances 2, 3, 7, and x

Rectangle ABCD divided through P for the Dual Pythagorean Theorem

x^2-7^2=b^2-a^2=3^2-2^2

x^2-49=9-4=5

x^2=54

x=\sqrt{54}=3\sqrt6

MA=MB \Longleftrightarrow MA^2-MB^2=0 \Longleftrightarrow PA^2-PB^2=0 \Longrightarrow PA=PB

Point P on the perpendicular bisector of AB

If two points $P,Q$ satisfy

PB^2-PA^2=QB^2-QA^2

then $P$ and $Q$ lie on a line perpendicular to $AB$

Points P and Q on a line perpendicular to AB

Assume on the contrary that $PQ$ is not perpendicular to $AB$

Draw two perpendiculars from $P,Q$ to $AB$

Perpendiculars from P and Q to AB meeting at C and D

DB^2-DA^2 = QB^2-QA^2 = PB^2-PA^2 = CB^2-CA^2

So

DB^2-DA^2=CB^2-CA^2

(DB-DA)(DB+DA) = (CB-CA)(CB+CA)

(DB-DA)AB=(CB-CA)AB

DB-DA=CB-CA

DB-DC-CA=CD+DB-CA

-DC=CD

2CD=0

CD=0 \quad \times

This gives a useful characterization:

\boxed{PB^2-PA^2=\text{constant}}

describes a line perpendicular to $AB$

Core

AMC 10A/12A 2025 (Problem 23/16)
AMC 10A/12A 2022 (Problems 23/20)

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