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Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset 91影视 Explanations$ Math

Student Edition 路 227 pages

ISBN: 9780395977279

Chapter 5: Q25 (page 182)

Given: Parallel planes P, Q and Rcutting transversal $\overset{鈫�}{AC}$ and $\overset{鈫�}{DF}$ ; $AB = BC$ .
Prove: $DE = EF$ .
(Hint: You can鈥檛 assume that $\overset{鈫�}{AC}$ and $\overset{鈫�}{DF}$ are coplanar. Draw $\overset{炉}{AF}$ , cutting plane Q at X. using the plane $\overset{炉}{AC}$ and $\overset{炉}{AF}$ , apply Theorems 3-1 and 5-10. Then use the plane of $\overset{炉}{AF}$ and $\overset{炉}{FD}$ .)

Short Answer

Expert verified

$DE = EF$

Step by step solution

01

Step 1. Consider the diagram.

The diagram is:

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

And $A B = B C$ .

02

Step 2. Show the proof.

In $螖 A B X$ and $螖 A C F$ ,

$鈭� A B X = 鈭� A C F$ $(B X 鈭� C F)$ .

$鈭� A X B = 鈭� A F C$ $(B X 鈭� C F)$ .

$鈭� A = 鈭� A$ (Common).

By AAA rule,

$螖 A B X ~ 螖 A C F$ .

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,

$X F = A X$ 鈥︹€� (1)

Now in triangle AFD and XFE:

In $螖 A F D$ and $螖 X F E$ ,

$鈭� F E X = 鈭� F D A$ $(X E 鈭� A D)$ .

$鈭� F X E = 鈭� F A D$ $(X E 鈭� A D)$

$鈭� F = 鈭� F$ (Common).

By AAA rule,

$螖 A F D ~ 螖 X F E$ .

So,

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

So,

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

By equation (1),

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

03

Step 3. State the conclusion.

Therefore, $D E = E F$ (proved).

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Most popular questions from this chapter

State the principal definition or theorem that enables you to deduce, from the information given, that quadrilateral SACK is a parallelogram.

$\overset{炉}{S A} 鈮� \overset{炉}{K C}; \overset{炉}{S A} 鈭� \overset{炉}{K C}$

M, N and T are the midpoints of the sides of $螖 XYZ$ .

If $TN = 7$ , then $XY = \underset{炉}{?}$

For exercises, 14-18 write paragraph proofs.

Given: parallelogram ABCD, $\overset{炉}{AN}$ bisects $鈭� DAB$ ; $\overset{炉}{CM}$ bisects $鈭� BCD$ .

Prove: AMCN is a parallelogram.

For exercises, 14-18 write paragraph proofs.

Given: parallelogram ABCD; W, X, Y, Z are midpoints of $\overset{炉}{AO}$ , $\overset{炉}{BO}$ , $\overset{炉}{CO}$ and $\overset{炉}{DO}$ .

Prove: role="math" localid="1637745814874" $WXYZ$ is a parallelogram.

$鈻� DECK$ is a parallelogram. Complete.

If $m 鈭� 1 = 3 x, m 鈭� 2 = 4 x$ and $m 鈭� 3 = x^{2} - 70$ then $x = ?$ and $m 鈭� CED = ?$ (numerical answers).

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