Question

In a parallelogram ABCD, M and N are the points on AB and BC respectively.

Prove that:

1. ΔCMD and ΔAND are equal in area. 
2. area (ΔAND) = area (ΔAMD) + area (ΔCMB).

Answer 1

Given:

ABCD is a parallelogram.

M is a point on AB and N is a point on BC.

To Prove:

1. area (ΔCMD) = area (ΔAND). 
2. area (ΔAND) = area (ΔAMD) + area (ΔCMB).

Proof [Step-wise]:

1. Set up coordinates (convenient choice).

Put A at the origin: A = (0, 0).

Let vector AB = b and AD = d, so B = b, D = d and C = b + d.

Let M divide AB so M = tb for some t (0 ≤ t ≤ 1).

Let N divide BC so N = B + sd = b + sd for some s (0 ≤ s ≤ 1). 

We will use the standard area formula for a triangle:

`"Area" (ΔPQR) = 1/2 |det (Q - P, R - P)|`

2. Prove (i): area (ΔCMD) = area (ΔAND).

Compute area (ΔCMD): 

`"Area" (ΔCMD) = 1/2 |det (M - C, D - C)|` 

= `1/2 |det (tb - (b + d), d - (b + d))|`

= `1/2 |det ((t - 1)b - d, -b)|` 

= `1/2 |det (-d, -b)|`

The term (t – 1)b gives zero when paired with `-b = 1/2 |det(d, b)|`.

Compute area (ΔAND):

`"Area" (ΔAND) = 1/2 |det (N - A, D - A)|`

= `1/2 |det (b + sd, d)|`

= `1/2 |det (b, d) + s xx det (d, d)|`

= `1/2 |det (b, d)|`   ...(Since det (d, d) = 0)

Note that |det (d, b)| = |det (b, d)|, so the two areas are equal. 

Therefore, area (ΔCMD) = area (ΔAND).

3. Prove (ii): area (ΔAND) = area (ΔAMD) + area (ΔCMB).

Compute area (ΔAMD): 

`"Area" (ΔAMD) = 1/2 |det (M - A, D - A)|` 

= `1/2 |det (tb, d)|` 

= `t xx 1/2 |det (b, d)|`

Compute area (ΔCMB):

`"Area" (ΔCMB) = 1/2 |det(M - C, B - C)|`

= `1/2 |det((t - 1)b - d, -d)|`

= `1/2 |(1 - t) det (b, d)|` 

= `(1 - t) xx 1/2 |det (b, d)|`

Sum: area (ΔAMD) + area (ΔCMB) 

= `t xx 1/2 |det (b, d)| + (1 - t) xx 1/2 |det (b, d)|` 

= `1/2 |det (b, d)|`

But from step 2 `"area" (ΔAND) = 1/2 |det (b, d)|`. 

Hence, area (ΔAND) = area (ΔAMD) + area (ΔCMB).