What can we say about the sum of the angles of any triangle? (Consider a triangle ABC and construct a line through A that is parallel to BC.)
Answer
The sum is always 180° — for every triangle, big or small, thin or fat.
The straight angle at A is split into ∠XAB, ∠A and ∠YAC — copies of ∠B, ∠A and ∠C.
XY ∥ BC → ∠B = ∠XAB and ∠C = ∠YAC (alternate angles) So ∠A + ∠B + ∠C = ∠A + ∠XAB + ∠YAC = 180°, as together they form a straight angle
This result is called the angle sum property of triangles.
Why it happens: The single clever step is drawing a line through the top vertex parallel to the base. It moves ∠B and ∠C up beside ∠A, where all three lie along one straight line. This idea appears in The Elements, the famous book of the Greek mathematician Euclid, who lived around 300 BCE.
Q2.
There is a convenient way of verifying the angle sum property by folding a triangular cut-out of a paper. Do you see how this shows that the sum of the angles in this triangle is 180°?
Answer
Yes — fold the three corners inwards and they line up along the base.
Cut out any paper triangle ABC and mark its three angles with different colours.
Fold the top vertex A down so that it lands exactly on the base BC.
Fold the corner at B inwards until it touches the same point, and do the same with the corner at C.
The three coloured corners now sit side by side, with no gap and no overlap, along the straight edge BC.
Three angles fitted along a straight line Straight angle = 180° So ∠A + ∠B + ∠C = 180°
Why it happens: The first fold is really the parallel line of the proof — the crease is parallel to BC. Folding turns the printed proof into something you can hold in your hand, which is why it convinces so quickly.
Try This: Repeat with a very obtuse triangle and with a right-angled one. The corners still fit exactly along the straight edge.
Q3.
Find ∠ACD, if ∠A = 50°, and ∠B = 60°.
Answer
∠ACD = 110°.
BC is extended to D; ∠ACD is the exterior angle at C.
From the angle sum property, 50° + 60° + ∠ACB = 180° 110° + ∠ACB = 180° So ∠ACB = 70°
∠ACB and ∠ACD together form a straight angle ∠ACD = 180° – 70° = 110°
Tip: An exterior angle is the angle between the extension of one side and the other side meeting it. Every triangle has six of them, two at each vertex.
Q4.
Find the exterior angle for different measures of ∠A and ∠B. Do you see any relation between the exterior angle and these two angles? [Hint: From angle sum property, we have ∠A + ∠B + ∠ACB = 180°.] We also have ∠ACD + ∠ACB = 180°, since they form a straight angle. What does this show?
Answer
The exterior angle always equals the sum of the two opposite interior angles.
∠A
∠B
∠ACB
∠ACD
∠A + ∠B
50°
60°
70°
110°
110°
40°
80°
60°
120°
120°
90°
30°
60°
120°
120°
35°
35°
110°
70°
70°
The two hints prove it in one line:
∠A + ∠B + ∠ACB = 180° … (angle sum property) ∠ACD + ∠ACB = 180° … (straight angle) Both right sides are 180°, so the left sides are equal: ∠A + ∠B + ∠ACB = ∠ACD + ∠ACB Take away ∠ACB from both sides: ∠ACD = ∠A + ∠B
Why it happens: Both expressions describe the same 180°, once through the triangle and once along the straight line. The angle ∠ACB is shared by both, so cancelling it leaves the exterior angle property: an exterior angle of a triangle equals the sum of its two remote interior angles.