External Angle Greater (Euclid 1-16)

For any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles.ABC is a triangle.

Consider side BC produced to D.

To Prove.

External angle ACD is greater than each of the internal and opposite angles, CBA and BAC.

PROOF

Let E be the mid-point of AC

Join BE and extend to F such that BE=EF

Join FC

Extend AC to G

AE = EC (Construct)

BE = EF (construct)

∠AEB = ∠FEC (Opp. ang.s of intersecting lines)

AB = CF

∠BAE = ∠FCE

∠ABE = ∠CFE

But

∠ACD > ∠FCE

∠ACD > ∠CAB

Similarly, by having cut BC in half, it can be shown that angle BCG is also greater than ABC.

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