Chapter 7.4, Problem 23WE

To determine

To prove: $NM\cdot CM=LM\cdot BM$ .

Explanation of Solution

Given:

The triangles are shown below:

  

Angles B and C are equal.

Calculation:

In triangle BMN and CML,

  $\begin{array}{l}\angle B=\angle C\left(\text{Given}\right)\\ \angle BMN=\angle CML\left(\text{Vertical opposite angles}\right)\\ \therefore \Delta BMN\sim \Delta CML\left(AA\right)\end{array}$

Hence:

  $\begin{array}{l}\frac{NM}{LM}=\frac{BM}{CM}\\ NM\cdot CM=LM\cdot BM\end{array}$

Hence, this proves that $NM\cdot CM=LM\cdot BM$ .