Question

In ΔABC, D and E are the mid-points of AC and AB, respectively. BD and CE produced meet a line through A parallel to BC at M and N, respectively. Prove that AM = AN.

Answer 1

Given: In triangle ABC, D and E are the mid-points of AC and AB respectively. BD and CE produced meet a line through A parallel to BC at M and N respectively.

To Prove: AM = AN

Proof [Step-wise]:

1. Choose a convenient coordinate system:

Put B = (0, 0), C = (2, 0) and A = (x, y) with y ≠ 0.

Choosing BC = 2 is only a scale choice and does not affect equality of lengths.

2. Since E is the midpoint of AB.

`E = ((0 + x)/2, (0 + y)/2)`

= `(x/2, y/2)`

Since D is the midpoint of AC.

`D = ((x + 2)/2, (y + 0)/2)`

= `((x+2)/2, y/2)` 

These are midpoint facts; see the midpoint theorem.

3. Parametrize the line BD:

Points on BD are t × D for t ∈ R because B is the origin. 

So, a general point on BD has coordinates `(t xx (x + 2)/2, t xx y/2)`.

The line through A parallel to BC is horizontal y = y.

To find M = BD ∩ line through A parallel to BC set the y-coordinate equal to y:

`t xx (y/2) = y`

⇒ t = 2

4. With t = 2 we get

M = 2D

= (x + 2, y)

Hence, the vector AM = M – A

= (x + 2 – x, y – y)

= (2, 0)

So, AM = 2.

5. Parametrize the line CE:

Points on CE are C + s(E – C).

Compute `E - C = (x/2 - 2, y/2 - 0)`.

So, a general point is `(2 + s(x/2 - 2), s xx y/2)`. 

Intersect with the horizontal line y = y by solving

`s xx y/2 = y`

⇒ s = 2

6. With s = 2 we get

N = C + 2(E – C)

= 2E – C

= (x – 2, y)

Hence, the vector AN = N – A

= (x – 2 – x, y – y)

= (–2, 0)

So, AN = 2.

7. Therefore, AM = 2 = AN, so AM = AN.