Geometry with Trigonometry

70 The parallel axiom; Euclidean geometry Ch. 5

CONVERSEofPythagoras’ theorem.Let A,B,C be non-collinear points such

that

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

Then∠BAC is a right-angle.
Proof. Choose the pointEso that|A,C|=|A,E|,Eis on the same side ofABas
Cis, and∠BAEis a right-angle. By Pythagoras’ theorem,

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

Thus|B,E|=|B,C|, and the lengths of the sides of the triangle[B,A,C]are equal to

those of[B,A,E]. By the SSS principle,[B,A,C]≡[B,A,E]. In particular|∠BAC|◦=

|∠BAE|◦and this latter is a right-angle by construction. In factE=C.

NOTE. In a right-angled triangle, the side opposite the right-angle is known as the

hypotenuse.

5.5 Mid-lines and triangles

5.5.1 Harmonicranges............................

Let A,B,C be non-collinear points such that|A,B|>|A,C|.TakeD=A so that A∈

[B,D]. Then the mid-lines of|BACand|CADmeet BC at points E,F, respectively,

such that{E,F}divide[B,C]internally and externally in the same ratio.

A

B C

H

EF

G

D

K

Figure 5.10.

Proof. By the cross-bar theorem the mid-line of|BACmeets[B,C]in a pointE.

LetGbe a point of the mid-line of|CAD,onthesamesideofABasCis. We cannot

haveAG‖BCas that would imply

|∠BCA|◦=|∠CAG|◦=|∠GAD|◦=|∠CBA|◦,

andthisinturnwouldimplythat|A,B|=|A,C|, contrary to hypothesis. ThenAG

meetsBCin some pointF.

TakeH∈[A,Dso that|A,H|=|A,C|.Then|∠AHC|◦=|∠ACH|◦.Wehavethat

|∠BAC|◦=|∠AHC|◦+|∠ACH|◦, |∠AHC|◦=|∠ACH|◦,

|∠BAC|◦=|∠BAE|◦+|∠EAC|◦, |∠BAE|◦=|∠EAC|◦.