Sec. 4.2 Alternate angles, parallel lines 53
CASE 2. LetB∈[E,C].Then[A,B ⊂IR(|DAC)and[D,B∈IR(|ADC).It
follows that
|∠BAC|◦=|∠DAC|◦−|∠DAB|◦=|∠ADC|◦−|∠ADB|◦=|∠BDC|◦.
CASE 3. LetC∈[B,E].Then[A,C⊂IR(|BAD)and[D,C
∈IR(|BDA). It follows that
|∠BAC|◦=|∠BAD|◦−|∠DAC|◦=|∠BDA|◦−|∠ADC|◦=|∠BDC|◦.
Now combining the cases, by the SAS principle, as|A,B|=|D,B|,|A,C|=|D,C|,|∠BAC|◦=|∠BDC|◦,
we haveT(A,B,C)→≡(D,B,C)T′′.ButT′′(D,B,C)→≡(A′,B′,C′)T′so
T(A,B,C)→≡(A′,B′,C′)T′.
This is known as theSSS(side, side,side)principle of congruence for triangles.4.2 Alternateangles,parallellines.....................
4.2.1 Alternateangles
Let A,B,C be non-collinear points, and take D=C so that C∈[A,D].Then
|∠BCD|◦>|∠CBA|◦.
Proof.LetE=mp(B,C)and chooseFso thatE=mp(A,F).ThenifT 1 ,T 2 are
the triangles with vertices{A,B,E},{F,C,E}, respectively, by the SAS principle of
congruenceT (^1) (A,B,E)→≡(F,C,E)T 2. In particular,
|∠EBA|◦=|∠ECF|◦,i.e.|∠CBA|◦=|∠BCF|◦.
But[C,F ⊂IR(|BCD)asE,andsoF, is in the closed half-plane with edgeACin
whichBlies, andDandFare on the opposite side ofBCfromA.AlsoF∈ADas
F∈ADwould imply thatE=C. Then by 3.5.2|∠BCF|◦<|∠BCD|◦.
COROLLARY 1.In the theorem let G=C be such that C∈[B,G].Then|∠ACG|◦
|∠ABC|◦.
Proof. This follows immediately as∠ACGand∠BCDare vertically opposite
angles.