Chapter 8.4, Problem 21PSB

Solution

a.

To prove that two triangles are similar.

Given: $\overline{AD}\parallel \overline{BC}$ , $AB=24,BC=9,AD=16$ and $DB=12$ are provided in the question which will help to find the solution.

Concept used: Here, we are using Side Angle similarity rule.

Proof: In $\Delta ABD$ and $\Delta BDC$

  $\angle CBD=\angle BDA$ (alternate interior angles)

And, $\frac{BC}{BD}=\frac{AD}{AB}\left(\because \frac{9}{12}=\frac{12}{16}\right)$

So, $\Delta ABD\sim \Delta BDC$ (by Side Angle similarity)

b.

The angle which is congruent to $\angle A$

  $\angle C$ is congruent to $\angle A$

Given: $\overline{AD}\parallel \overline{BC}$ , $AB=24,BC=9,AD=16$ and $DB=12$ are provided in the question which will help to find the solution.

Concept used: Here, we are using the concept of corresponding angles.

Calculation: Here, $\angle A=\angle C$ (corresponding angles)

Hence, we can say that

  $\angle A\cong \angle C$

Conclusion: $\angle C$ is congruent to $\angle A$

c.

The measure of CD

The measure of CD is $13.5$

Given: $\overline{AD}\parallel \overline{BC}$ , $AB=24,BC=9,AD=16$ and $DB=12$ are provided in the question which will help to find the solution.

Calculation: Here, we know that

  $\Delta ABD\sim \Delta BDC$ (proved above)

So, $\frac{CD}{BA}=\frac{BC}{AD}$

➔ $\frac{CD}{24}=\frac{9}{16}$

➔ $CD=\frac{27}{2}=13.5$

Conclusion: The measure of CD is $13.5$