Prove that any two sides of a triangle are together greater than twice the median drawn to third side.
To prove that any two sides of a triangle are always greater than twice the median drawn to the third side, let's consider the triangle ABC, where AB, BC, and AC represent the sides of the triangle, and AD is the median drawn to side BC.
First, let's assume that BC is the longest side of the triangle, so AC > BC and AB > BC.
According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In our case, applying this theorem to triangle ABC, we get:
AC + AB > BC -- (1)
AB + BC > AC -- (2)
BC + AC > AB -- (3)
Now, let's consider the triangle ABD, where AD represents the median drawn to side BC.
Using the Triangle Inequality Theorem again, we can state:
AD + AB > BD -- (4)
AD + BD > AB -- (5)
BD + AB > AD -- (6)
Since AD is a median, we know that BD = CD (as medians bisect the opposite side).
Combining (1) and (4), we have:
AC + AB > BC
AD + AB > BD
Adding these two inequalities together, we get:
(AC + AB) + (AD + AB) > BC + BD
AC + 2AB + AD > BC + BD -- (7)
Since AC = BC + CD (according to the Triangle Inequality Theorem), we can substitute this into equation (7):
BC + CD + 2AB + AD > BC + BD
Simplifying:
BC + 2AB + AD > BC + BD
We can cancel out BC from both sides, leaving us with:
2AB + AD > BD -- (8)
Since BD = CD (as mentioned before), we can rewrite inequality (8) as:
2AB + AD > CD
Thus, we have shown that any two sides of a triangle (AB and AC) are always greater than twice the median AD drawn to the third side BC.
The same proof can be applied for the cases when AB or AC are the longest sides of the triangle.