Question

use the given information to complete the proof of the following theorem. if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally. given: $overline{ed}paralleloverline{gf}$. prove: $\frac{eg}{gc}=\frac{df}{fc}$. statement reason 1. $overline{ed}paralleloverline{gf}$ given 2. $angle cedcongangle cgf$ reason? 3. $angle ccongangle c$ reflexive property 4. $\triangle cdesim\triangle cfg$ reason? 5. $\frac{ec}{gc}=\frac{dc}{fc}$ corr. sides of similar triangles are proportional 6. $ec = eg + gc$ reason? 7. $dc = df+fc$ reason? 8. $\frac{eg + gc}{gc}=\frac{df + fc}{fc}$ substitution property 9. $\frac{eg}{gc}+\frac{gc}{gc}=\frac{df}{fc}+\frac{fc}{fc}$ fraction algebra 10. $\frac{eg}{gc}+1=\frac{df}{fc}+1$ fraction algebra 11. $\frac{eg}{gc}=\frac{df}{fc}$ reason?

Explanation:

Step1: Identify corresponding - angles

Since $\overline{ED}\parallel\overline{GF}$, by the corresponding - angles postulate, $\angle CED\cong\angle CGF$.

Step2: Use reflexive property

$\angle C\cong\angle C$ by the reflexive property of congruence.

Step3: Prove similar triangles

By the AA (angle - angle) similarity criterion, $\triangle CDE\sim\triangle CFG$ because we have two pairs of congruent angles ($\angle CED\cong\angle CGF$ and $\angle C\cong\angle C$).

Step4: Write proportion of similar - triangle sides

For similar triangles $\triangle CDE$ and $\triangle CFG$, the corresponding sides are proportional, so $\frac{EC}{GC}=\frac{DC}{FC}$.

Step5: Segment - addition postulate

By the segment - addition postulate, $EC = EG+GC$ and $DC = DF + FC$.

Step6: Substitute segment lengths

Substitute $EC = EG + GC$ and $DC = DF+FC$ into the proportion $\frac{EC}{GC}=\frac{DC}{FC}$, we get $\frac{EG + GC}{GC}=\frac{DF + FC}{FC}$.

Step7: Apply fraction algebra

Using fraction algebra, $\frac{EG + GC}{GC}=\frac{EG}{GC}+\frac{GC}{GC}$ and $\frac{DF + FC}{FC}=\frac{DF}{FC}+\frac{FC}{FC}$. Since $\frac{GC}{GC} = 1$ and $\frac{FC}{FC}=1$, we have $\frac{EG}{GC}+1=\frac{DF}{FC}+1$.

Step8: Subtract 1 from both sides

Subtract 1 from both sides of the equation $\frac{EG}{GC}+1=\frac{DF}{FC}+1$. By the subtraction property of equality, $\frac{EG}{GC}=\frac{DF}{FC}$.

Answer:

The reasons are as follows:

1. Corresponding - angles postulate
2. $C$; Reflexive property of congruence
3. AA similarity criterion
4. $DC$; Corresponding sides of similar triangles are proportional
5. Segment - addition postulate
6. Segment - addition postulate
7. $DF$; $FC$
8. Subtraction property of equality