Everything covered in this lesson, in one place - useful for revision or printing.
\(\triangle ABC\)
A triangle has 3 sides and 3 interior angles. Every triangle has exactly 180° worth of angles inside — no matter its shape or size. This lesson covers all the properties you need.
\(\text{Scalene} \quad AB \neq BC \neq CA\)
A scalene triangle has all three sides of different lengths, so all three angles are also different.
\(\text{Isosceles} \quad DE = DF\)
An isosceles triangle has exactly two equal sides. The angles opposite those equal sides (the base angles) are also equal.
\(\text{Equilateral} \quad PQ = QR = RP\)
An equilateral triangle has all three sides equal, so all three angles are also equal.
\(\text{Acute} \quad \hat{A} < 90°,\; \hat{B} < 90°,\; \hat{C} < 90°\)
An acute triangle has all three interior angles less than 90°.
\(\text{Right-angled} \quad \hat{A} = 90°\)
A right-angled triangle has exactly one angle equal to 90°. The side opposite the right angle is called the hypotenuse.
\(\begin{gathered}\text{Obtuse} \quad \hat{A} > 90°\\[6pt]\textbf{3 by sides, 3 by angles}\end{gathered}\)
An obtuse triangle has one angle greater than 90°. You now know all six triangle types. A triangle can belong to one type from each group — e.g. an isosceles right-angled triangle.
\(\hat{A} + \hat{B} + \hat{C} = 180°\)
The three interior angles of any triangle always add up to 180°. This is true for every triangle ever drawn.
\((4x+10)° + (5x+5)° + 2x° = 180°\)
The three angles of triangle ABC are \((4x+10)°\), \((5x+5)°\) and \(2x°\). Use the angle sum to solve for \(x\).
\(11x + 15 = 180\)
Collect like terms: \(4x + 5x + 2x = 11x\) and \(10 + 5 = 15\). So \(11x + 15 = 180\).
\(11x = 165 \implies x = 15\)
Subtract 15: \(11x = 165\). Divide by 11: \(x = 15\).
\(\begin{gathered}x = 15,\quad \hat{A}=70°,\;\hat{B}=80°,\;\hat{C}=30°\\[4pt]\hat{A}+\hat{B}+\hat{C}=180°\end{gathered}\)
Angle Sum Theorem: the three interior angles of any triangle add to 180°. Always set up the equation and solve for the unknown.
\(DE = DF \implies \hat{E} = \hat{F}\)
In an isosceles triangle the two base angles (opposite the equal sides) are always equal.
\(\hat{G} = 22°,\quad \hat{J} = 22°\)
Triangle GHJ has \(\hat{G} = \hat{J} = 22°\) (two equal angles → isosceles). Find \(\hat{H}\).
\(\hat{H} = 180° - 22° - 22° = 136°\)
Use the angle sum: \(\hat{H} = 180° - 22° - 22° = 136°\).
\(\begin{gathered}\hat{H}=136°\\[4pt]\hat{E}=\hat{F}\implies \text{isosceles}\end{gathered}\)
Isosceles property: two equal sides → two equal base angles. If any two angles in a triangle are equal, it is isosceles.
\(PQ = QR = RP\)
All three sides are equal in an equilateral triangle.
\(\hat{P} = \hat{Q} = \hat{R}\)
Because all sides are equal, all three angles must also be equal.
\(\begin{gathered}\hat{P}=\hat{Q}=\hat{R}=60°\\[4pt]\frac{180°}{3}=60°\end{gathered}\)
Equilateral triangle: all sides equal → all angles equal → each angle is exactly 60°. Always.
\(\text{Extend side } BC \text{ to point } D\)
When you extend one side of a triangle, the angle formed outside the triangle at that vertex is called an exterior angle.
\(\hat{ACD}_{\text{ext}} = \hat{A} + \hat{B}\)
The exterior angle equals the sum of the two non-adjacent interior angles (the two angles that are not next to it). This follows directly from the angle sum theorem.
\(x_\text{ext} = 55° + 72° = 127°\)
If two interior angles are 55° and 72°, the exterior angle at the third vertex is \(55°+72°=127°\). Check: the interior angle there is \(180°-127°=53°\) and \(55°+72°+53°=180°\)
\(\begin{gathered}\hat{ACD}=\hat{A}+\hat{B}\\[4pt]\text{exterior} = \text{sum of remote interior angles}\end{gathered}\)
Exterior Angle Theorem: an exterior angle of a triangle equals the sum of the two non-adjacent (remote) interior angles.
\(\hat{A} = 90°\)
In a right-angled triangle one angle is exactly 90°. The side opposite the right angle is the hypotenuse — always the longest side.
\(\hat{B} + \hat{C} = 90°\)
Because the three angles sum to 180° and one is already 90°, the other two must sum to 90°. We call them complementary angles.
\(\hat{B}=41°,\;\hat{A}=90° \implies \hat{C}=49°\)
Given \(\hat{B}=41°\) and \(\hat{A}=90°\): \(\hat{C}=180°-90°-41°=49°\). Or simply: \(\hat{C}=90°-41°=49°\).
\(\begin{gathered}\hat{A}=90°\\[4pt]\hat{B}+\hat{C}=90°\end{gathered}\)
Right triangle rule: the two non-right angles are complementary (add to 90°). Knowing any one of them gives you the other.
\(AD \perp BC \implies \angle ADB = \angle ADC = 90°\)
An altitude is a perpendicular line drawn from a vertex to the opposite side (or its extension). It forms two right angles at the base.
\(\hat{A}=90°,\;\hat{B}=41° \implies y = \angle BAD = 49°\)
If \(AD \perp BC\) so \(\hat{ADB}=90°\) and \(\hat{B}=41°\), then in triangle ABD: \(y = 180°-90°-41°=49°\).
\(MW \perp NV \text{ and } NW = WV\)
A perpendicular bisector cuts a side at 90° and at its exact midpoint. Two things at once: right angle + equal halves.
\(\begin{gathered}\text{Altitude: vertex} \perp \text{opposite side}\\[4pt]\text{Perp. bisector: midpoint} + 90°\end{gathered}\)
Altitude: from vertex, forms 90° at base. Perpendicular bisector: cuts a side at its midpoint at 90°. When a triangle has a perpendicular bisector from the apex, the two base segments are equal.
\(\begin{gathered}\hat{A}+\hat{B}+\hat{C}=180°\\[4pt]\text{Isosceles: } \hat{E}=\hat{F}\\[4pt]\text{Equilateral: each}=60°\\[4pt]\text{Exterior}=\hat{A}+\hat{B}\\[4pt]\text{Right: }\hat{B}+\hat{C}=90°\end{gathered}\)
You have covered all 7 topics. These 5 key properties appear in almost every triangle problem. Now test yourself with the quiz!