Question

Given: `DeltaABC` with `bar(DE)` || `bar(AC)` 

Answer 1

Given: ΔABC with DE║AC

To Prove: $\frac{AD}{DB} =\frac{CE}{EB}$

Statements and Reasons

1. DE║AC (given)

2. ∡CAB = ∡EDB and ∡ACB = ∡DEB (If two parallel lines are cut by a transversal, the corresponding angles are congruent.
)

3. ΔABC ≈ ΔDBE (Angle-Angle Similarity theorem)

4. $\frac{AB}{DB} =\frac{CB}{EB}$ (Corresponding sides of similar triangles are proportional.)

5. AB = AD + DB
and CB = CE + EB
(segment addition postulate)

6. $\frac{AD + DB}{DB} =\frac{CE + EB}{EB}$ (Substitution Property of Equality
)

7. $\frac{AD}{DB}\; + 1 =\; \frac{CE}{EB}\; + 1$
(Division property)

8. $\frac{AD}{DB} =\frac{CE}{EB}$ (Subtraction Property of Equality
)

So missing statement in the proof was $\frac{AB}{DB} =\frac{CB}{EB}$

Answer 2

Answer:

$\frac{\text{AB}}{\text{DB}}=\frac{\text{CB}}{\text{EB}}$

Step-by-step explanation:

Since AC || DE, and ABC forms a triangle, we can deduce that ABC is a  smaller triangle inside triangle DBE.

In the second step, we see that ∠CAB in the smaller triangle is congruent to ∠EDB in the larger one, due to the corresponding angles theorem.  We also see that ∠ACB in the smaller triangle is congruent to ∠DEB in the larger one, again by the corresponding angles theorem.

This leads to the third step; since we have two of the angles in one triangle congruent to the corresponding angles in the other triangle, this means the third angle will be congruent (since all triangles equal 180°, and the first two angles are equal, the third must be equal as well).  Two triangles in which all three angles of one triangle are congruent to the three angles of the other are similar triangles.

Since the triangles are similar, corresponding sides have the same ratio.  This means that the ratio of AB to the larger side DB is the same as the ratio of CB to the larger side EB; this gives us

AB/DB = CB/EB.