**Proof Through Geometry**

The figure shows an equilateral triangle ABC with DE parallel to BC and EF parallel to AB. D and G are on AB; F and H are on BC. GM and HN are perpendicular to AC. If AC = a, DG = d, FH = f, and MN = x prove that x = (a + d + f) / 2

Equilateral Triangle ABC

AC = a

DE // BC, EF // AB

DG = d

FH = f

GM perpendicular to AC

HN perpendicular to AC

**Given:**Equilateral Triangle ABC

AC = a

DE // BC, EF // AB

DG = d

FH = f

GM perpendicular to AC

HN perpendicular to AC

**Prove:**

x=(a+d+f)/2## Hint 1: DE // BC, EF // AB

By having parallel sides with our large equilateral triangle, ABC, we notice that triangle ADE and triangle EFC are also equilateral triangles. Therefore AD = DE = AE and

EF = FC = EC.

EF = FC = EC.

## Hint 2: Other Than a Triangle

Notice the parallelogram and use it to find more equal lines.

Therefore AE = DE = BF and EC = EF = BD.

Therefore AE = DE = BF and EC = EF = BD.

## Hint 3: Hint 3: Start with *a*

in *a+d+f*

Substitute a in for the line segments it represents from the triangle.

*a*= AE + EC## Hint 4: Create Midpoint *k *on *a*

There is now a line BK that splits the equilateral triangle in half.

We now have two 30-60-90 triangles. With these triangles come

special properties such as the hypotenus is twice as much as the base.

Therefore AB = 2AK and BC = 2KC

We now have two 30-60-90 triangles. With these triangles come

special properties such as the hypotenus is twice as much as the base.

Therefore AB = 2AK and BC = 2KC

## Hint 5: Mess With the *x*

Now

Therefore BG = 2GT and BH = 2VH. Use this information to substitute in the eaquarions and information gathered.

*x*= MK + KN = GT + VH and still holding the same properties as the previos hint.Therefore BG = 2GT and BH = 2VH. Use this information to substitute in the eaquarions and information gathered.

## Solution

1. By looking at the triangles we can say that a = AE + EC

2. a + d + f = AE + EC + d + f

3. Replace AE with equivalent BF, and EC with DB and you can

now measure two line segments.

4. BF + DB + d + f = GB + BH

5. Make midpoint K. By making this midpoint we have two 30-60-90

triangles.

6. We can draw two new line segments such that we're measuring

x using these 30-60-90 triangles.

7. GB + BH = a + d + f

8. x = MK + KN but we can now use our new values x = GT + VH

9. GB = 2 GT and BH = 2VH

10. Substituting values in for step 7 we get: a + d + f = 2GT + 2VH

11. a + d + f = 2(GT + VH)

12. a + d + f = 2(x)

13. Dividing both sides by 2: x = (a + d + f) / 2

2. a + d + f = AE + EC + d + f

3. Replace AE with equivalent BF, and EC with DB and you can

now measure two line segments.

4. BF + DB + d + f = GB + BH

5. Make midpoint K. By making this midpoint we have two 30-60-90

triangles.

6. We can draw two new line segments such that we're measuring

x using these 30-60-90 triangles.

7. GB + BH = a + d + f

8. x = MK + KN but we can now use our new values x = GT + VH

9. GB = 2 GT and BH = 2VH

10. Substituting values in for step 7 we get: a + d + f = 2GT + 2VH

11. a + d + f = 2(GT + VH)

12. a + d + f = 2(x)

13. Dividing both sides by 2: x = (a + d + f) / 2