Chapter 8.5, Problem 40SPR

To determine

To prove:

  $\frac{AD}{DB}=\frac{AC}{BC}$

Explanation of Solution

Given information:

  

  $\overline{EH}\parallel \overline{FG},\overline{EF}\parallel \overline{HG}$ , $\overline{JF}$ bisects $\angle EFG$

Proof:

  

The given information is:

  $\overline{EH}\parallel \overline{FG},\overline{EF}\parallel \overline{HG}$ , $\overline{JF}$ Bisects $\angle EFG$

By using triangle proportionality theorem:

  $\frac{EK}{KF}=\frac{GJ}{JF}\mathrm{...}\left(1\right)$

Then define the angle bisector:

  $\angle EFK\cong \angle KFG$

Corresponding angle postulate:

  $\angle KFG\cong \angle JKH$

The theorem of vertical angle:

  $\angle JKH\cong \angle EKF$

The transitive property:

  $\angle EFK\cong \angle EKF$

Alternate interior angle theorem:

  $\angle FJH\cong \angle EKF$

Similarity theorem of AA:

  $\Delta EKF\sim \Delta GJF$

Similar polygon theorem:

  $\frac{EK}{KF}=\frac{GJ}{JF}$

Hence, this condition is proved.