Inequalities

Theorem:

Statement: If two sides of a triangle are unequal, the angle opposite to the longer side is greater than the angle opposite to the shorter side.

Proof:

Let $\triangle ABC$ have $AC > AB$. We must prove that $\angle ABC > \angle ACB$.

Construction: Since $AC > AB$, take a point $D$ on $AC$ such that $AD = AB$. Join $BD$.

1. In $\triangle ABD$, since $AD = AB$, by Pons Asinorum: $\angle ABD = \angle ADB$   --- (1)
2. Now, in $\triangle BDC$, $\angle ADB$ is an exterior angle for vertex $D$. By the Exterior Angle Theorem: an exterior angle of a triangle is strictly greater than either interior opposite angle: $$\angle ADB > \angle DCB \implies \angle ADB > \angle ACB \quad \text{--- (2)}$$
3. From (1) and (2): $\angle ABD > \angle ACB$.
4. Clearly, $\angle ABC > \angle ABD$ (since $BD$ lies inside $\angle ABC$).
5. Therefore, $\mathbf{\angle ABC > \angle ACB}$. $\blacksquare$

Theorem:

Statement: The sum of any two sides of a triangle is strictly greater than the third side.

For any triangle $\triangle ABC$ with sides $a, b, c$:

Corollary (Difference of Sides):

Subtracting $b$ from both sides of $a + b > c$ gives $a > c - b$, or more generally:

The absolute difference between any two sides of a triangle is strictly less than the third side.

Condition for Third Side Range:

Given two sides $a$ and $b$ ($a \ge b$), the third side $c$ must strictly satisfy the bounding inequality:

Theorem:

Statement: Of all line segments that can be drawn to a given line from a given point not lying on it, the perpendicular line segment is the shortest.

Proof:

Let $l$ be a line, and let $P$ be a point not on $l$. Draw $PM \perp l$ (meeting $l$ at $M$). Let $N$ be any other point on $l$ distinct from $M$. Join $PN$.

In $\triangle PMN$, $\angle M = 90^\circ$.

Since the sum of angles of a triangle is $180^\circ$ and $\angle M = 90^\circ$, $\angle N$ must be an acute angle ($< 90^\circ$).

By the side-angle inequality theorem, the side opposite to the greater angle is longer:

Since $N$ was an arbitrary point on line $l$, $PM$ is strictly the shortest distance from $P$ to line $l$. $\blacksquare$

Problem 1: Can a triangle have side lengths $4\text{ cm}, 7\text{ cm}, 12\text{ cm}$? Justify your answer.

Solution:

Check the Triangle Inequality Theorem: The sum of the two smaller sides must exceed the largest side.

Since $11\text{ cm} < 12\text{ cm}$ (the sum is strictly less than the third side), no such triangle can exist.

Problem 2: Prove that the perimeter of a triangle is greater than the sum of its three medians.

Solution:

Let $AD$ be the median to side $BC$ of $\triangle ABC$. Produce $AD$ to $E$ such that $DE = AD$. Join $EC$.

In $\triangle ADB$ and $\triangle EDC$:

1. $BD = CD$ (Since $D$ is mid-point of $BC$).
2. $\angle ADB = \angle EDC$ (Vertically opposite angles).
3. $AD = ED$ (By construction).

By SAS: $\triangle ADB \cong \triangle EDC \implies AB = EC$.

Now, in $\triangle AEC$, by the Triangle Inequality Theorem:

Substitute $EC = AB$ and $AE = 2AD$:

Similarly, for medians $BE$ and $CF$:

Adding inequalities (1), (2), and (3):