Question

In the adjoining figure, AB || FD, AC || GE and BD = EC. Prove that:

1. BG = DF
2. EG = CF.

Answer 1

Given:

AB || FD, AC || GE and BD = EC in triangle ABC with D and E on BC, F on AC and G on AB as in the figure.

To Prove:

1. BG = DF
2. EG = CF

Proof [Step-wise]:

1. Place coordinates: Let B = (0, 0), C = (c, 0) (c > 0). 

Let A = (0, a) with a > 0.

This choice makes AB vertical; the argument below is purely affine and equivalent to a synthetic proof using parallel-lines properties.

2. Put E = (e, 0) point on BC. 

Since BD = EC and B is at 0. 

D is at x = d where BD = d and EC = c – e.

BD = EC 

⇒ d = c – e

So, D = (d, 0) = (c – e, 0).

3. Find G intersection of the line through E parallel to AC with AB:

Slope of AC

= `(0 - a)/(c - 0)`

= `-a/c` 

A line through E = (e, 0) with that slope has equation `y = (-a/c)(x - e)`.

Its intersection with AB (x = 0) gives G = `(0, a e/c)`. 

Hence BG = distance from B(0, 0) to `G(0, a e/c) = a e/c`.

Using the standard alternate-angle / parallel-line facts to get the same proportional relation is allowed; see parallel-line theorems.

4. Find F intersection of the line through D parallel to AB with AC.

Because AB is vertical, the line through D parallel to AB is vertical x = d.

Its intersection with AC line from A(0, a) to C(c, 0) with equation `y = a(1 - x/c)` is `F = (d, a(1 - d/c))`.

So DF = vertical distance = `a(1 - d/c)`.

5. Use d = c – e to compare BG and DF:

`BG = a e/c`, 

`DF = a(1 - d/c)`

= `a(1 - (c - e)/c)`

= `a(e/c)` 

Therefore, BG = DF.

This proves (i).

6. Compute EG and CF to prove (ii).

EG = distance between E(e, 0) and `G(0, a e/c)`:

`EG = sqrt(e^2 + (a e/c)^2)`

= `e xx sqrt(1 + (a^2/c^2))`

CF = distance between C(c, 0) and `F(d, a(1 - d/c))`.

Note c – d

= c – (c – e) 

= e 

And `1 - d/c = e/c` 

So, the vector CF

= `(d - c, a(1 - d/c))` 

= `(-e, a e/c)`

Hence, `CF = sqrt(e^2 + (a e/c)^2)`

= `e xx sqrt(1 + (a^2/c^2))`

Thus, EG = CF.

This proves (ii).

(i) BG = DF and (ii) EG = CF, as required.