Question

In the adjoining figure, ABCD is a parallelogram. A line through A intersects DC at P and meets BC produced at Q.

Prove that : area (ΔBCP) = area (ΔDPQ).

Answer 1

Given: ABCD is a parallelogram. A line through A meets DC at P and meets BC produced at Q figure as given.

To Prove: area (ΔBCP) = area (ΔDPQ).

Proof [Step-wise]:

1. Put a convenient coordinate system:

Take A = (0, 0), AB = (1, 0) so B = (1, 0).

 Let AD = (x, y) (y ≠ 0), so D = (x, y) and C = B + D = (1 + x, y).

This places DC horizontal at the y-coordinate y.

2. Let the line through A have slope m; its equation is y = mx.

Its intersection with DC the horizontal line y = y gives P: 

m × x_P = y 

⇒ `x_P = y/m` 

So, `P = (y/m, y)`.

3. The line BC is the line through B in direction d = AD = (x, y); a parametric point is B + t × d = (1 + tx, ty).

Intersecting this with y = mx the line through A gives

t × y = m(1 + tx) 

⇒ `t = m/(y - mx)`

Thus, Q = (1 + tx, ty)

= `(y/(y - mx), (my)/(y - mx))`

4. Use the vector determinant formula for area:

For triangle with vertices U, V, W, area (ΔUVW) 

= `1/2 |det (V - U, W - U) |`

For ΔBCP: Take U = B, V = C, W = P.

Then `C - B = (x, y), P - B = (y/m - 1, y)`.

`det (C - B, P - B) = x xx y - y xx (y/m - 1)` 

= `xy - (y^2/m) + y`

Hence, `"area" (ΔBCP) = 1/2 xx |xy - y^2/m + y|`.

5. For ΔDPQ: Take U = D, V = P, W = Q.

Then `P - D = (y/m - x, 0)`.

Since P_y = D_y = y .

`Q - D = (y/(y - mx) - x, (my)/(y - mx) - y)` 

The determinant simplifies because P – D has zero second component:

`det (P - D, Q - D) = (y/(m - x)) xx ((my)/(y - m x) - y)`

Simplify the product:

`(y/(m - x)) xx ((my)/(y - mx) - y)` 

= `((y - mx)/m) xx (y(m/(y - mx) - 1))`

= `(y/m) xx (m - y + mx)`

= `xy - y^2/(m + y)`

Thus, `"area" (ΔDPQ) = 1/2 xx |xy - y^2/(m + y)|`.

6. From steps 4 and 5, the two area expressions are equal, so area (ΔBCP) = area (ΔDPQ).

Therefore, area (ΔBCP) = area (ΔDPQ), as required.