Source — Euclid's Elements, encoded as an rrxiv paper

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paper/figures/fig-i-32.pngimage · 8904 bytesRaw

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(I.11), with

C E = A C = C B

. Join

E A

and

E B

; through

D

draw

D F

parallel to

C E

(I.31), meeting

E B

at

F

; through

F

draw

F G

parallel to

A B

(I.31), meeting

C E

at

G

. Join

A F

. Since

∠ E C A

is a right angle and

C A = C E

, the triangle

A C E

is right-isoceles, and

∠ C A E = ∠ A E C =

half a right angle (I.5; I.32). Similarly

∠ C B E = ∠ C E B =

half a right angle. Hence

∠ A E B

is a right angle (Common Notion 2), and triangle

A E B

is right-angled at

E

. By I.47: \[ AB^2 \;=\; AE^2 + EB^2. \] Now

A E^{2} = A C^{2} + C E^{2} = 2 \cdot A C^{2}

(I.47 in

△ A C E

, plus

C E = A C

). Similarly inside the right triangles formed by the perpendicular

D F

at

D

on

A B

, I.47 plus the fact that

D F = D E

(which one shows from the parallel construction) yields, after Common Notions 2 and 3: \[ AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2. \]