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, we can use the property that the area of a triangle is half the product of its base and height.
First, let's consider the area of triangle BDF. The base of triangle BDF is BD.
To find the height of triangle BDF, draw a perpendicular from E to BD (let's call the point of intersection G). Since BD is parallel to AC (given that ABCD is a parallelogram), we have AG = CE as corresponding sides of similar triangles.
Since CF = AE (given in the question), we know that AG + GD = CE + GD = CD. So, AG = CD - GD.
Now, the height of triangle BDF is BG, which is equal to AG because it is a perpendicular drawn from the base of a triangle. So, the height of triangle BDF is CD - GD.
Therefore, the area of triangle BDF is:
Area of BDF = (1/2) * base * height
= (1/2) * BD * (CD - GD)
Now, let's consider the area of parallelogram ABCD. The base of parallelogram ABCD is also BD, and the height is the distance between BD and AC. Since AC is produced to F, we can use the fact that CF = AE to find this distance.
Let's call the distance between BD and AC as h.
Since CF = AE, we have CD = CE + ED = CF + ED = CF + DF. So, DF = CD - CF.
Now, the height of parallelogram ABCD is h, which is equal to DF because it is a perpendicular drawn from the base of the parallelogram. So, the height of parallelogram ABCD is CD - CF.
Therefore, the area of parallelogram ABCD is:
Area of ABCD = base * height
= BD * (CD - CF)
To prove that Area of BDF = Area of ABCD, we need to show that the two expressions are equal.
(1/2) * BD * (CD - GD) = BD * (CD - CF)
To establish this equality, we can cancel out BD from both sides of the equation:
(1/2) * (CD - GD) = CD - CF
Simplifying the left side:
(1/2) * CD - (1/2) * GD = CD - CF
Moving all terms to one side:
(1/2) * CD - CD + (1/2) * GD = -CF
Combining the like terms:
(1/2) * GD - (1/2) * CD = -CF
Multiplying both sides by 2:
GD - CD = -2 * CF
Now, we know that CF = AE. So, substituting this into the equation:
GD - CD = -2 * AE
Rearranging the terms:
GD + 2 * AE = CD
But we also know that CD = GD + DF. So, substituting this:
GD + 2 * AE = GD + DF
Cancelling out GD from both sides:
2 * AE = DF
And we know that DF = h, the height of parallelogram ABCD.
So, 2 * AE = h
Now, we can substitute this result back into the equation we obtained earlier:
(1/2) * (CD - GD) = CD - CF
(1/2) * CD - (1/2) * GD = CD - CF
(1/2) * CD - GD = CD - AE
(1/2) * CD - GD = CD - (1/2) * h
Multiplying both sides by 2:
CD - 2 * GD = 2 * CD - h
Rearranging the terms:
2 * GD = CD + h
But we know that CD = GD + DF, so substituting this:
2 * GD = GD + DF + h
Cancelling out GD from both sides:
GD = DF + h
And we know that GD = BG, the height of triangle BDF.
So, BG = DF + h
And from earlier, we found that 2 * AE = h.
Substituting this:
BG = DF + 2 * AE
Therefore, the heights of triangle BDF and parallelogram ABCD are equal.
Since the base and the height are equal, the area of triangle BDF is equal to the area of parallelogram ABCD.
Hence, we have proved that ar(BDF) = ar(parallelogram ABCD).