Chapter 7.6, Problem 2CE

To determine

To complete: The given statements

Answer to Problem 2CE

  $\frac{lowerleft}{wholeleft}=\frac{lowerright}{wholeright}$

  $\frac{upperleft}{lowerleft}=\frac{upperright}{lowerright}$

  $\frac{upperleft}{wholeleft}=\frac{upperparallel}{lowerparallel}=\frac{upperright}{wholeright}$

Explanation of Solution

  

According to Basic proportionality theorem, if a line intersects two sides at distinct points and parallel to the third side then the other two sides are divided in the same proportion.

As $DE\parallel BC$ ,

  $\frac{AD}{BD}=\frac{AE}{EC}$ (i)

  $\begin{array}{l}\Rightarrow \frac{AD}{BD}+1=\frac{AE}{FC}+1\\ \Rightarrow \frac{AD+BD}{BD}=\frac{AE+EC}{EC}\\ \Rightarrow \frac{AB}{BD}=\frac{AC}{EC}\\ \Rightarrow \frac{BD}{AB}=\frac{CE}{AC}(ii)\end{array}$

Now prove $\Delta ADE\sim \Delta ABC$

Now $DE\parallel BC$ . Also, take AB as a transversal.

  $\angle ADE=\angle ABC$

(Each pair of alternate interior angles are equal when a transversal intersects two parallel lines)

Also, $DE\parallel BC$ such that take AC as a transversal.

  $\angle AED=\angle ACB$

(Each pair of alternate interior angles are equal when a transversal intersects two parallel lines)

Therefore,

  $\Delta ADE\sim \Delta ABC$

(according to AA similarity criteria, if two angles of one triangle are equal to the two angles of another triangle then the triangles are similar)

So,

  $\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{EC}(iii)$

(If two triangles are similar then the following conditions are satisfied:

1. Corresponding sides of the polygons are proportional.

2. Corresponding angles of the polygons are equal.)

In the given figure,

Lower left is BD

Lower right is BD

Whole left is AB

Whole right is AC

Upper left is AD

Upper right is AE

Upper parallel is DE

Lower parallel is BC

From (i),

  $\frac{upperleft}{lowerleft}=\frac{upperright}{lowerright}$

From (ii),

  $\frac{lowerleft}{wholeleft}=\frac{lowerright}{wholeright}$

From (iii),

  $\frac{upperleft}{wholeleft}=\frac{upperparallel}{lowerparallel}=\frac{upperright}{wholeright}$