Chapter 8.4, Problem 12PSB

To determine

Prove $AB\cdot FD=HB\cdot GF$ .

Answer to Problem 12PSB

Proved $AB\cdot FD=HB\cdot GF$

Explanation of Solution

Given information:

  

Parallelogram $ACEG$ with $F$ the midpoint of $EG$ , $\angle ABH\cong \angle EFD$ .

Calculation:

  $ACEG$ is a Parallelogram

Opposite angles of parallelogram are congruent,

  $\angle HAB\cong \angle DEF$

  $\angle ABH\cong \angle EFD$ (given)

  $\Delta HAB\cong \Delta DEF$

If two triangles are similar then corresponding side length are proportional,

  $\frac{AB}{FE}=\frac{HB}{DF}$

  $F$ is the midpoint of $EG$

Midpoint divides the segment in to two congruent parts,

  $\begin{array}{l}GFiscongruentEF\\ \frac{AB}{GF}=\frac{HB}{DF}\left[Transitiveproperty\right]\\ AB\cdot FD=HB\cdot GF\left[extreamproducttheorem\right]\end{array}$

Hence, Proved $AB\cdot FD=HB\cdot GF$