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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)

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  • $\begingroup$ What does make two angles congruent? $\endgroup$ – Berci Mar 11 at 1:31
  • $\begingroup$ Hint.. If $a + b = c + b$ what can you say about how the numbers $a$ and $c$ are related? Hint: What is the definition of suplimentary angles? $\endgroup$ – fleablood Mar 11 at 1:35
  • $\begingroup$ That's overly complicated! Why are you inventing so many unnecessary things. You have four angles. You don't need anything else. We asked you two simple questions. 1) What makes two angles congruent? and 2) What is the definition of supplementary? Answer those. The prove should be easy. Almost trivial. $\endgroup$ – fleablood Mar 11 at 15:18
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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:

$\angle C \cong \angle D$.

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