Geometry with Trigonometry

Sec. 5.4 Pythagoras’ theorem, c. 550B.C. 69

Now the line throughB′′which is parallel toBCwill meet[A,C in a pointDsuch
that
|A,B′′|
|A,B|

=|A,D|

|A,C|

Hence
|A,C′′|
|A,C|

=

|A,D|

|A,C|

,

from which it follows that|A,D|=|A,C′′|.AsC′′,D∈[A,Cwe then haveD=C′′and
soB′′C′′‖BC. Thus the degree-measures of the angles of[A,B,C]are equal to those
of the corresponding angles of[A,B′′,C′′]and so in turn to those of the corresponding
angles in[A′,B′,C′].

5.4 Pythagoras’ theorem, c. 550B.C.

5.4.1

Pythagoras’ theorem.Let A,B,C be non-collinear points such that AB⊥AC. Then

|B,C|^2 =|C,A|^2 +|A,B|^2.

Proof.LetDbe the foot of the perpendicular fromAtoBC; then by 4.3.3Dis be-
tweenBandC. The triangles[D,B,A],[A,B,C]are similar as|∠ADB|◦=|∠CAB|◦=
90 ,|∠DBA|◦=|∠ABC|◦, and then by 5.2.2|∠BAD|◦=|∠BCA|◦. Then by the last
result
|A,B|
|B,C|

=

|B,D|

|A,B|

,

so that|A,B|^2 =|B,D||B,C|. By a similar argument applied to the triangles[D,C,A],
[A,B,C]we get that|A,C|^2 =|D,C||B,C|. Then by addition, asD∈[B,C],

|A,B|^2 +|A,C|^2 =(|B,D|+|D,C|)|B,C|=|B,C|^2.

A

BD C

Figure 5.8. Pythagoras’ theorem.

B A

C E

Figure 5.9. Impossible figure

for converse.