prove that if the lengths of two sides of a and b, triangle are a and b respectively, then the lengths of the corresponding altitudes to those sides are in the ratio b/a.
To prove that the lengths of the corresponding altitudes to the sides of a triangle are in the ratio b/a, we can use the concept of similar triangles.
Let's represent our original triangle as ABC, where side BC has length a and side AC has length b. Let's also denote the corresponding altitudes as h1 and h2, where h1 is the altitude corresponding to side BC, and h2 is the altitude corresponding to side AC.
To begin the proof, we construct a new triangle XYZ, where side XZ has length b and side YZ has length a. We draw the perpendiculars, h1' and h2', from vertices Y and Z to sides XZ and YZ, respectively.
Now, we have two triangles, ABC and XYZ. It can be observed that both triangles are similar, as they have the same angles. This is because the angles in a triangle are purely determined by its shape and do not change even if the lengths of the sides change.
From the similarity of triangles ABC and XYZ, we can establish the following ratio:
h1'/h2' = a/b
This ratio is derived from the property of similar triangles, which states that the ratios of corresponding sides of similar triangles are equal.
Now, we compare the ratios h1'/h2' and h1/h2. Note that the two pairs of altitudes, h1' and h1, as well as h2' and h2, are corresponding altitudes to the respective sides of the triangles ABC and XYZ.
Since h1'/h2' = a/b, and we want to prove that h1/h2 = b/a, we need to somehow flip the ratio a/b to b/a.
To do this, we take the reciprocal of both sides of the first equation:
1/(h1'/h2') = 1/(a/b)
This gives us:
h2'/h1' = b/a
Now, let's consider the ratios h2'/h1' and h2/h1. Both of these ratios represent the corresponding altitudes to the sides of triangles ABC and XYZ.
Since h2'/h1' = b/a, and we want to prove that h2/h1 = b/a, we just need to observe that the ratios h2'/h1' and h2/h1 are equal, because both ratios represent the same quantity (the ratio of corresponding altitudes to corresponding sides of the triangles).
Therefore, we conclude that h2/h1 = b/a, proving that the lengths of the corresponding altitudes to the sides of a triangle are in the ratio b/a.
To prove that the lengths of the corresponding altitudes to the sides of a triangle are in the ratio b/a, we can use the area as a basis.
Let's consider a triangle ABC with side lengths a, b, and c. Assume that h₁ and h₂ are the corresponding altitudes to sides a and b, respectively.
The area of a triangle can be expressed as A = (1/2) * base * height. Therefore, we can write:
Area of triangle ABC = (1/2) * a * h₁ ...(1)
Area of triangle ABC = (1/2) * b * h₂ ...(2)
Dividing equation (1) by equation (2), we get:
(a * h₁) / (b * h₂) = 1
Now, let's consider the ratio of the side lengths a/b:
a/b = (Area of triangle ABC with side length a) / (Area of triangle ABC with side length b)
Since the base of the triangle is unchanged (it is equal to c in both cases), we can express the ratio as:
a/b = (a * h₁) / (b * h₂)
Cross multiplying the equation, we have:
a * h₁ = b * h₂
Dividing both sides by a and b respectively, we get:
h₁ / h₂ = b / a
Therefore, we have proved that the lengths of the corresponding altitudes to the sides of a triangle are in the ratio b/a.