Note: Your teacher will grade your response to this question to ensure you receive proper credit for your answer. Complete the proof. Given: AB · BE = CB · BD Prove: ΔABC ~ ΔDBE
To prove that ΔABC is similar to ΔDBE, we need to show that their corresponding angles are congruent and their corresponding sides are proportional.
1. Given: AB · BE = CB · BD
2. As AB · BE = CB · BD, we can rewrite it as (AB/BD) = (CB/BE).
3. Let's extend AB and CB to form a straight line.
Then, let's draw a line segment from D to intersect AC at point F.
4. By the Intercept Theorem, we have AF/FD = CB/BD.
5. Since (AB/BD) = (CB/BE) = (AF/FD) = k (let k be the constant of proportionality), AB/BD = k and CB/BE = k.
6. In ΔABC and ΔDBE, we have:
- ∠ABC = ∠DBE (vertical angles formed by the intersecting lines AB and BE)
- ∠ACB = ∠DEB (vertical angles formed by the intersecting lines CB and BD)
- AB/BD = k (established in step 5)
- CB/BE = k (established in step 5)
7. Thus, by the Angle-Angle Similarity Postulate (AAA), we can conclude that ΔABC ~ ΔDBE.