Question

In the adjoining figure, PQ || BC, BE || CA and CF || BA. Prove that area of ΔABE = area of ΔACF.

Answer 1

Given: PQ || BC, BE || CA and CF || BA. Points E and F lie on a line EF, which contains P and Q, so EF contains segment PQ, hence EF || BC.

To Prove: Area (ΔABE) = Area (ΔACF).

Proof [Step-wise]:

1. Place a coordinate system to simplify computation (valid synthetic technique).

Put B = (0, 0), C = (1, 0) so BC is the x-axis.

Let A = (0, h) with h ≠ 0 (A above the base).

Because PQ ⊥? (given PQ || BC) the line EF that contains P and Q is parallel to BC.

So, EF is a horizontal line y = k for some k (0 < k < h in the picture).

2. AC has slope `(0 - h)/(1 - 0) = -h`. 

Since BE || AC and passes through B(0, 0), the equation of BE is y = –hx.

Its intersection with EF (y = k) gives E:

k = –h_xE 

⇒ `x_E = -k/h`

So, `E = (-k/h, k)`.

3. AB is the vertical line x = 0.

Since CF || AB and passes through C(1, 0), CF is the vertical line x = 1.

Its intersection with EF (y = k) gives F = (1, k).

4. Compute area (ΔABE).

Using the determinant formula for area.

Area (ΔABE) = `1/2` |det(B – A, E – A)

B – A = (0, 0) – (0, h)

= (0, –h)

`E - A = (-k/h, k) - (0, h)` 

= `(-k/h, k - h)`

`det = (0)(k - h) - (-h)(-k/h)`

= 0 – k

= –k

Hence, Area (ΔABE) = `1/2 |-k| = k/2`.

5. Compute area (ΔACF).

C – A = (1, 0) – (0, h)

= (1, –h)

F – A = (1, k) – (0, h)

= (1, k – h)

det = (1)(k – h) – (–h)(1)

= k – h + h

= k

Hence, Area (ΔACF) = `1/2 |k| = k/2`.

6. From steps 4 and 5, we get

`"Area" (ΔABE) = k/2 = "Area" (ΔACF)`

Area (ΔABE) = Area (ΔACF).

Hence proved.