Question

Given: BD bisects ABC. Auxiliary EA is drawn such that AE || BD. Auxiliary BE is an extension of BC

Prove: AD/DC = AB/BC

Could you please fill out picture 2? I'm horrible with this type of math.

Answer 1

Reasons 2 and following

2. Definition of angle bisector

3. Given

4. They are corresponding angles

5. Transitive property of equality

6. Alternate angles are congruent

7. Transitive property of equality

8. From 7 ΔABE is isosceles

9. Definition of isosceles triangle

10. Triangle proportionality theorem

11. Substituting 9 in 10

Answer 2

The required proof is given in the table below:


`\begin{tabular}{|p{4cm}|p{6cm}|} Statement & Reason \\ [1ex] 1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\ 2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\ 5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 6. \angle ABD\cong\angle BAE & 6. Alternate angles \end{tabular}`

`\begin{tabular}{|p{4cm}|p{6cm}|} 7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\ 8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\ 9. EB = AB & 9. De(finition of congruence \\ 10. $(AD)/(DC)=(EB)/(BC)$ & 10. Triangle proportionality theorem \\ 11. $(AD)/(DC)=(AB)/(BC)$ & 11. Substitution Property of equality \\[1ex] \end{tabular}`