Proposition 3.14: Geometry

Question

Please help prove the following proposition: Proposition 3.14 Supplementary angles of congruent angles are congruent.

Is this right? (1) Suppose angle ABC is congruent to angle DEF (given) (2) We have arbitrary points A, C, and G on the sides of angle ABC, and the supplement angle CBG of angle ABC. We can choose points D, F, and H such that AB≅DE,CB≅FE,and BG≅EH. (C-1) (3) Triangle ABC is congruent to triangle DEF (C-6) (4) So AC≅DF and ∡A≅∡D. (def cong triangles) (5) Also AG≅DH (C-3) (6) So triangle ACG is congruent to triangle DFH (C-6 SAS) (7) So CG≅FH and angle G ≅angle H (def cong triangles) (8) So triangle CBD is cong to triangle FEH (C-6 SAS) (9) Then angle CBG is congruent to angle FEH (def cong triangles)

Answer 1

That's overly complicated! Why are you inventing so many unnecessary things? You have four angles. You don't need anything else.

Given: $\angle A \cong \angle B$. $\angle C$ is supplementary to $\angle A$. $\angle D$ is supplementary to $\angle B$.

Pf: 1)$m\angle A = m\angle B$ (def of congruent)

2)$m\angle A + m \angle C = 180$ (def of supplementary)

3)$m\angle D + m \angle B = 180$ (def of supplementary)

The rest is algebra:

In 3) replace $m \angle B$ with $m\angle A$ to get:

$m\angle D + m \angle A = 180$

Compare $m\angle A + m \angle C = 180$ with $m\angle D + m \angle A = 180$ to get:

$m\angle A + m \angle C=m\angle D + m \angle A$

Subtract $m\angle A$ from both sides to get:

$m\angle C = m\angle D$ and conclude:

$\angle C \cong \angle D$.