Chapter 5: Q25 (page 182)

Given: Parallel planes P, Q and Rcutting transversal $\overset{\leftrightarrow}{AC}$ and $\overset{\leftrightarrow}{DF}$ ; $AB = BC$ .
Prove: $DE = EF$ .
(Hint: You can’t assume that $\overset{\leftrightarrow}{AC}$ and $\overset{\leftrightarrow}{DF}$ are coplanar. Draw $\bar{AF}$ , cutting plane Q at X. using the plane $\bar{AC}$ and $\bar{AF}$ , apply Theorems 3-1 and 5-10. Then use the plane of $\bar{AF}$ and $\bar{FD}$ .)

Short Answer

Expert verified

$DE = EF$

Step by step solution

Step 1. Consider the diagram.

The diagram is:

Here Parallel planes P, Q and R cutting transversal $\overset{\leftrightarrow}{A C}$ and $\overset{\leftrightarrow}{D F}$ ;

And $A B = B C$ .

Step 2. Show the proof.

In $Δ A B X$ and $Δ A C F$ ,

$∠ A B X = ∠ A C F$ $(B X ∥ C F)$ .

$∠ A X B = ∠ A F C$ $(B X ∥ C F)$ .

$∠ A = ∠ A$ (Common).

By AAA rule,

$Δ A B X ~ Δ A C F$ .

So,

$\frac{A B}{A C} = \frac{B X}{C F} = \frac{A X}{A F}$

So,

$\begin{matrix} \frac{A B}{A B + B C} = \frac{A X}{A X + X F} \\ \frac{A X + X F}{A X} = \frac{A B + B C}{A B} \\ 1 + \frac{X F}{A X} = 1 + \frac{B C}{A B} \\ \frac{X F}{A X} = \frac{B C}{A B} \end{matrix}$

Since, $A B = B C$ , then,

$X F = A X$ …… (1)

Now in triangle AFD and XFE:

In $Δ A F D$ and $Δ X F E$ ,

$∠ F E X = ∠ F D A$ $(X E ∥ A D)$ .

$∠ F X E = ∠ F A D$ $(X E ∥ A D)$

$∠ F = ∠ F$ (Common).

By AAA rule,

$Δ A F D ~ Δ X F E$ .

So,

$\frac{F A}{F X} = \frac{F D}{F E} = \frac{X E}{A D}$ .

So,

$\begin{matrix} \frac{F X + A X}{F X} = \frac{F E + E D}{F E} \\ 1 + \frac{A X}{F X} = 1 + \frac{F D}{F E} \\ \frac{A X}{F X} = \frac{E D}{F E} \end{matrix}$