ABCD ia a parallelogram whose diagonals intersecting at E. AC is produced to F such that CF=AE. Prove that ar(BDF)=ar(parallelogram ABCD)

To prove that the area of triangle BDF is equal to the area of parallelogram ABCD, let's analyze the given information step by step.

1. Given: ABCD is a parallelogram.
2. Diagonals AC and BD intersect at E.
3. AC is produced to F such that CF = AE.

To prove that the area of triangle BDF is equal to the area of parallelogram ABCD, we need to prove that the triangles BDF and ABC have the same base and equal heights.

Let's break it down into two separate parts:

Part 1: Triangles BDF and ABC have the same base.

Since ABCD is a parallelogram, its opposite sides are parallel.
Therefore, we have BD || AC.

Since CF = AE and AC is extended to F, we can conclude that the line segment CF is the extension of diagonal AC.

Now, let's consider the triangles BDF and ABC:

In triangle BDF:
- The base is BD.

In triangle ABC:
- The base is AC.

Since BD || AC, the base of triangle BDF (BD) is parallel to the base of triangle ABC (AC).

Therefore, triangles BDF and ABC have the same base.

Part 2: Triangles BDF and ABC have equal heights.

Let's consider the height of triangle ABC. The height of a parallelogram is defined as the perpendicular distance between its base and its opposite side.

Since ABCD is a parallelogram, the height of triangle ABC is the perpendicular distance between base AC and side BD.

Let's consider the height of triangle BDF. The height of a triangle is defined as the perpendicular distance between its base and its opposite vertex.

In triangle BDF, the height is the perpendicular distance between base BD and vertex F.

Now, let's analyze the given information:

1. Diagonals AC and BD intersect at E.
2. AC is extended to F such that CF = AE.

Since AC = BD (opposite sides of a parallelogram), and CF = AE (given), we can conclude that the line segment CF is equal in length to the height of triangle ABC.

Furthermore, the line segment CF is perpendicular to BD, which is the base of triangle BDF.

Therefore, the height of triangle BDF is equal to the height of triangle ABC.

Now, we have established that triangles BDF and ABC have the same base and equal heights.

By the formula for the area of a triangle (Area = 0.5 x base x height), we can conclude that the area of triangle BDF is equal to the area of triangle ABC.

Since triangle BDF is a part of parallelogram ABCD, it follows that the area of BDF is equal to the area of the entire parallelogram ABCD.

Hence, ar(BDF) = ar(parallelogram ABCD).

Therefore, we have proved that the area of triangle BDF is equal to the area of parallelogram ABCD.

To prove that the areas of triangle BDF and parallelogram ABCD are equal, we need to show that they have the same base and height.

Let's break down the proof step by step:

1. Draw the parallel lines: Draw lines AFC and BE such that they are parallel to CD.

2. Identify congruent triangles: In parallelogram ABCD, triangle CEA is congruent to triangle FCB (by ASA congruence). Therefore, all corresponding sides and angles are congruent.

3. Identify additional congruent triangles: Triangles CBF and ADE are congruent (by SAS congruence) because side CF is equal to side AE and angles CBF and ADE are congruent (both are opposite angles).

4. Determine the ratio of the areas: Since triangles CBF and ADE are congruent, their areas are equal. Therefore, the ratio of the area of triangle CBF to triangle ABC is equal to the ratio of the area of triangle ADE to triangle ABC.

5. Prove that triangles BDF and ABC have the same area: Since the ratio of the areas of triangles CBF and ABC is equal to the ratio of the areas of ADE and ABC, it follows that the ratio of the areas of triangles BDF and ABC is also equal.

6. Conclude: As the base of both triangles BDF and ABC is BD, and their heights are equal (they share the same height from the base BC), the areas of these two triangles are equal. Hence, ar(BDF) = ar(parallelogram ABCD).

This completes the proof.