in an equilateral triangle ABC,O is any point in the interior of the triangle.From
O perpendiculars are drawn to the sides.prove that sum of these perpendiculars is constant for any triangle
Viviani's Theorem proves this for isosceles triangles. If you take a look at
http://www.cut-the-knot.org/triangle/VivianiByVectors.shtml
where it is proved for equilateral triangles, you should be able to see that it applies equally well to arbitrary triangles.
If not, you may be able to find a suitable extension to Viviani's Theorem.
To prove that the sum of perpendiculars from any point inside an equilateral triangle to its sides is constant, we can utilize the property of similar triangles.
Let's consider an equilateral triangle ABC, and let O be any point inside the triangle. Let the perpendiculars from O intersect the sides at P, Q, and R as shown below:
A
/ \
/ \
/_____\
B C
\ /
\ /
\ /
O
/|\
/ | \
/ | \
P Q R
We need to prove that the sum of OP + OQ + OR is a constant value.
To start, let's draw lines AO, BO, and CO as shown:
A
/ \
/ \
/_____\
B C
\ /
\ /
\ /
O
/|\
/ | \
/ | \
P Q R
/ \ | / \
/ \ | / \
/_____\|/_____\
A' O' C'
| | Here, A', O', and C' are the midpoints of the respective sides.
Now, in triangle AOC and triangle A'O'C', we can see that they are similar triangles. This is because angle AOC is equal to the angle A'O'C' (both are right angles), and angle ACO is equal to angle A'C'O' (both are 90 degrees in an equilateral triangle).
Since the two triangles AOC and A'O'C' are similar, we can establish the following proportions:
OP/OA = O'P/O'A' (1)
OQ/OB = O'Q/O'B' (2)
OR/OC = O'R/O'C' (3)
Now, let's simplify the above proportions based on the properties of an equilateral triangle:
OA = OB = OC = a (side length of the equilateral triangle)
A'O' = B'O' = C'O' = a/2
OP = OQ = OR = h (height of the equilateral triangle - distance between any side and its opposite vertex)
Substituting these values into equations (1), (2), and (3), we get:
h/a = O'P/(a/2)
h/a = O'Q/(a/2)
h/a = O'R/(a/2)
Simplifying these equations, we get:
2O'P = h (4)
2O'Q = h (5)
2O'R = h (6)
Summing up equations (4), (5), and (6), we obtain:
2O'P + 2O'Q + 2O'R = 3h
Dividing both sides of the equation by 2, we get:
O'P + O'Q + O'R = h/2
Considering that O'P, O'Q, and O'R are the perpendiculars drawn from point O to the sides of the equilateral triangle, it means that the sum of these perpendiculars is equal to half of the height of the equilateral triangle, h/2.
Hence, the sum of the perpendiculars from any point O inside an equilateral triangle to its sides is constant and equal to half of the height of the equilateral triangle.