Question

In the adjoining figure, PQ || BC.

Prove that :

1. area (ΔABQ) = area (ΔАСР). 
2. area (ΔΒΟP) = area (ΔCOQ).

Answer 1

Given: In triangle ABC, points P on AB and Q on AC are such that PQ || BC. Lines BQ and CP meet at O. Figure as given.

To Prove:

1. area (ΔABQ) = area (ΔАСР). 
2. area (ΔΒΟP) = area (ΔCOQ).

Proof [Step-wise]:

I. Preliminary observation (similarity/proportionality)

1. Since PQ || BC, triangles APQ and ABC are similar. 

Hence, `(AP)/(AB) = (AQ)/(AC)`.   ...(Call this common ratio k)

II. Proof of (i): area (ΔABQ) = area (ΔACP)

1. Put vectors u = B − A and v = C − A. 

Then P = A + tu and Q = A + tv for the same t.   ...`(because (AP)/(AB) = (AQ)/(AC) = t = k)`

2. `"Area" (ΔABQ) = 1/2 |(B - A) xx (Q - A)|`

= `1/2 |u xx (tv)|`

= `t xx 1/2 |u xx v|`

3. `"Area" (ΔACP) = 1/2 |(C - A) xx (P - A)|`

= `1/2 |v xx (tu)|`

= `t xx 1/2 |v xx u|` 

= `t xx 1/2 |u xx v|`

4. Therefore, area (ΔABQ) = area (ΔACP).

III. Proof of (ii): area (ΔBOP) = area (ΔCOQ)

1. Choose coordinates B(0, 0), C(1, 0), A(0, 1).

Let the horizontal line through P and Q have y = p (0 < p < 1).

Then P = (0, p) on AB and Q = (1 – p, p) on AC, since PQ || BC (horizontal).

2. Parametrize BQ: 

BQ is (α(1 – p), αp)

Parametrize CP:

CP is (1 – β, βp) 

Their intersection O satisfies α = β and α(1 – p) = 1 – α.

So, `α = 1/(2 - p)`.

Hence, `O = ((1 - p)/(2 - p), p/(2 - p))`.

3. `"Area" (ΔBOP) = 1/2 xx |det (O - B, P - B )|`

= `1/2 xx x_O xx p`

= `1/2 xx p xx (1 - p)/(2 - p)`

4. `"Area" (ΔCOQ) = 1/2 xx |det (O - C, Q - C )|`

= `1/2 xx p xx (1 - p)/(2 - p)`   ...(Computation shown in stepwise algebra above)

5. Hence, area (ΔBOP) = area (ΔCOQ).