In the diagram, ABCD is a parallelogram with AB = 14 units. AD is produced to E such that AD:DE = 4:3. EB intersects DC in F. EB = 21 units. Calculate, with reasons, the length of FB.

Since ABCD is a parallelogram, AB is parallel to DC. This means AB is parallel to DF.
Consider $\triangle EDF$ and $\triangle EAB$:

• $\angle DEF = \angle AEB$ (Common angle)
• $\angle EDF = \angle EAB$ (Corresponding angles, since DF $\parallel$ AB and AE is a transversal)
• $\angle DFE = \angle ABE$ (Corresponding angles, since DF $\parallel$ AB and EB is a transversal) Therefore, $\triangle EDF \sim \triangle EAB$ (AAA similarity).

Given AD : DE = 4 : 3. Let $AD = 4x$ and $DE = 3x$.
Then $AE = AD + DE = 4x + 3x = 7x$.
The ratio of corresponding sides is $\frac{ED}{EA} = \frac{3x}{7x} = \frac{3}{7}$.
From the similarity, we have $\frac{EF}{EB} = \frac{ED}{EA}$.

• $\angle DEF = \angle AEB$ (Common angle)
• Since ABCD is a parallelogram, $AB \parallel DC$. Therefore, $AB \parallel DF$.
• $\angle EDF = \angle EAB$ (Corresponding angles, as $AB \parallel DF$ and AE is a transversal)
• $\angle DFE = \angle ABE$ (Corresponding angles, as $AB \parallel DF$ and EB is a transversal)

Since all three corresponding angles are equal, by the Angle-Angle-Angle (AAA) similarity criterion:

From the similarity $\triangle EDF \sim \triangle EAB$ (proven in 9.2), we have:

We know $\frac{ED}{EA} = \frac{3}{7}$ (from 9.1) and $AB = 14$ units.

Since ABCD is a parallelogram, opposite sides are equal in length.

Now, we can find FC: