If D and E are the midpoints of the sides AB and AC respectively of triangle ABC .F and G are the trisection of the base BC then prove that 3triangleFGH= triangleABC
To prove that 3triangleFGH = triangleABC, we can use the concept of similar triangles.
1. Let's start by drawing triangle ABC and marking the midpoints D and E on sides AB and AC respectively. Also, mark the points F and G, which divide the base BC into three equal parts.
2. Now, we can see that triangle ADE is similar to triangle ABC by the midpoint theorem. This means that the corresponding sides of these triangles are in proportion.
3. Similarly, triangle FGH is also similar to triangle ABC. This can be seen by considering that F and G are the trisection points of the base BC, so the segments BF, FG, and GC are all equal.
4. Using the similarity of triangle ADE and triangle ABC, we can write the following proportion:
AB/AD = AC/AE
5. Using the similarity of triangle FGH and triangle ABC, we can write the following proportion:
AB/FG = AC/FH
6. Since FG = GC and AC = AE + EC, we can express AC/AE as (AE + EC)/AE, which simplifies to 1 + EC/AE.
7. Similarly, FH = BG and AB = AD + DB, so AB/FG can be expressed as (AD + DB)/(AD + BG), which simplifies to 1 + BG/AD.
8. Combining steps 5, 6, and 7, we can write:
1 + BG/AD = 1 + EC/AE
9. Since the ratio of the magnitudes of the corresponding sides of similar triangles are equal, we can conclude that:
BG/AD = EC/AE
10. Now, multiplying both sides by 3, we have:
3(BG/AD) = 3(EC/AE)
11. This can be written as:
3(BG/AD) = 3(EC/AE) = 3(BC/AC)
12. Finally, we can see that 3(BG/AD) represents the ratio of the lengths of the corresponding sides of the triangles FGH and ABC. Similarly, 3(BC/AC) represents the ratio of the areas of the triangles FGH and ABC.
13. Therefore, we can conclude that 3triangleFGH = triangleABC.
This completes the proof that the area of triangle FGH is three times the area of triangle ABC.
To prove that 3*triangle FGH = triangle ABC, we need to show that the area of triangle FGH is one-third of the area of triangle ABC.
To do this, let's use the properties of midpoints and trisection points:
1. The midpoints D and E divide the sides AB and AC into two equal parts.
2. The trisection points F and G divide the base BC into three equal parts.
Now, let's break down the two triangles and compare their areas:
First, consider triangle ABC:
- The base BC has length x.
- The height from vertex A to base BC is h.
The area of triangle ABC is given by: (1/2) * x * h.
Next, consider triangle FGH:
- The base GH also has length x since F and G trisect BC.
- The height from vertex F to base GH is also h, as it is parallel to the height of triangle ABC.
The area of triangle FGH is given by: (1/2) * x * h.
Now, comparing the areas of the two triangles:
Area of triangle FGH / Area of triangle ABC = [(1/2) * x * h] / [(1/2) * x * h] = 1.
Since the ratio of the areas is 1, this means that triangle FGH has the same area as triangle ABC.
However, we want to prove that 3*triangle FGH = triangle ABC. To show this, we multiply both sides of the equation by 3:
3 * area of triangle FGH = 3 * area of triangle ABC.
Since the area of triangle FGH is equal to the area of triangle ABC, the equation becomes:
3 * area of triangle ABC = 3 * area of triangle ABC.
Thus, we have proved that 3*triangle FGH = triangle ABC.
no idea where H is
but clearly FGH is not congruent to ABC. Did you mean similar?
If so, look for congruent angles, or sides in a constant ratio.