Question

In a trapezium ABCD, AB || DC. A line parallel to AC cuts AB at P and BC at Q. Prove that area (ΔАPD) = area (ΔACQ).

Answer 1

Given:

ABCD is a trapezium with AB || DC.

A line through P on AB and Q on BC is drawn so that PQ || AC.

To Prove: area (ΔAPD) = area (ΔACQ).

Proof [Step-wise]:

1. Coordinate setup: Put A at (0, 0), AB on the x-axis and DC parallel to AB at y = –h (h > 0). 

Thus, take B = (b, 0), D = (0, –h), C = (c, –h).

Let P = (p, 0) be the given point on AB (0 ≤ p ≤ b).

This choice preserves AB || DC and places A, B on one horizontal line and D, C on a parallel horizontal line.

2. Area of ΔAPD: Points A(0, 0), P(p, 0), D(0, –h) give a triangle with base AP = p and height h.

Hence, area (ΔAPD)

= `1/2` × base × height

= `1/2` × p × h

3. Equation of AC and of the line through P parallel to AC:

AC has slope `-h/c`, so any line parallel to AC has equation `y = (-h/c)(x - s)` for some s.

The line through P(p, 0) is `y = (-h/c)(x - p)`.

4. Equation of BC: Line through B(b, 0) and C(c, –h) has slope `m = -h/(c - b)`. 

So, `BC: y = [-h/(c - b)](x - b)`.

5. Find Q as intersection of the two lines PQ and BC.

Equate `(-h/c)(x - p) = (-h/(c - b))(x - b)`.

Cancel –h (nonzero) and solve:

`(1/c)(x - p) = (1/(c - b))(x - b)`

⇒ (c – b)(x – p) = c(x – b) 

⇒ `x = c + p(1 - c/b)`

Call this x_Q.

6. Compute area of ΔACQ using A(0, 0), C(c, –h), Q(x_Q, y_Q).

With A at origin,

`"area" (ΔACQ) = 1/2 |x_C xx y_Q - x_Q xx y_C|` 

= `1/2 |c xx y_Q + h xx x_Q|`

Use y_Q from BC :

`y_Q = [-h/(c - b)](x_Q - b)`

Then `c xx y_Q + h xx x_Q = h xx b/(c - b) xx (c - x_Q)`.

7. Substitute x_Q found in step 5:

`c - x_Q = p xx (c - b)/b`

Hence, c × y_Q + h × x_Q 

= `h xx b/(c - b) xx p xx (c - b)/b`

= h × p

8. Therefore, area (ΔACQ) 

= `1/2 xx (h xx p)` 

= `1/2 xx p xx h`

9. Compare with step 2: 

area (ΔAPD)

= `1/2 xx p xx h` 

= area (ΔACQ)

area(ΔAPD) = area(ΔACQ), as required.