Chapter 4.5 — Exercise 4.4 — Angle Sum of Triangle

The Exterior Angle, Derived Rather Than Memorised

When side AC of triangle ABC is extended to D, ∠BCD (outside the triangle) and ∠BCA (inside it) form a linear pair, summing to 180°. But ∠A + ∠B + ∠BCA also sum to 180°, by the angle sum property. Since both sums equal the same 180°, and ∠BCA appears in both, subtracting it from each side leaves ∠BCD = ∠A + ∠B — the exterior angle theorem isn't a separate fact to memorise so much as these two 180° totals compared against each other.

Applying the Exterior Angle Theorem Directly

Given any two interior angles, the exterior angle at the remaining vertex is just their sum: 50° and 60° give an exterior angle of 110° there, with no need to first find the third interior angle and then subtract it from 180° as a roundabout alternative. Working backward, if the exterior angle and one interior angle are known, subtracting gives the other — an exterior angle of 135° alongside a known interior angle of 70° gives the remaining interior angle as 65°. Occasionally the "known" interior angle isn't given directly but has to be recovered first — a linear pair with some other marked angle, say — before the exterior angle theorem's subtraction step can even be attempted.

Some problems chain the theorem across two triangles: an exterior angle of one triangle becomes an interior angle feeding into a second triangle sharing a vertex with it. Solving the inner triangle first, then carrying that result into the outer triangle's own angle sum or exterior angle relationship, is the only way through — there's no shortcut that skips the inner triangle. For instance, given ∠PRT = 40° and ∠RPT = 95° as two angles of triangle PRT, the exterior angle at T (∠PTS, formed by extending RT) equals their sum: 40° + 95° = 135°. That same ∠PTS is also, in a second triangle sharing vertex T, an ordinary interior or exterior angle again — so whatever value it takes in the first triangle carries straight into the second one's own angle relationships, with no separate measurement needed.

Bisectors Inside a Triangle

In triangle XYZ, ∠X = 62° and ∠XYZ = 54°. First, the angle sum property gives ∠XZY = 180° − 62° − 54° = 64°. If YO and ZO bisect ∠XYZ and ∠XZY respectively, meeting at O, then ∠OYZ = 27° and ∠OZY = 32° — each simply half its parent angle. Applying the angle sum property once more, this time to the smaller triangle OYZ: ∠YOZ = 180° − 27° − 32° = 121°.

That second formula isn't a coincidence specific to this triangle — it holds generally, for the angle formed by the two bisectors of any triangle's other two angles. Since ∠OYZ + ∠OZY is always half of (∠XYZ + ∠XZY), which is itself (180° − ∠X), the angle at O works out to 180° minus half of (180° − ∠X), which simplifies to 90° + ∠X/2 — always obtuse, since ∠X/2 is always positive. This is a genuinely useful self-check when solving any bisector-meeting-point problem: if the computed angle at O ever comes out acute, something upstream in the working has gone wrong, since the general formula guarantees an obtuse result no matter which triangle the bisectors start from.

Mixing Parallel Lines With Triangles

When a problem gives both parallel lines and a triangle, the parallel-line properties usually need to run first, feeding their result into the triangle's angle sum afterward. With AB ∥ CD and BC ∥ DE: since AB ∥ CD, the angle 3x° and a given 105° angle are alternate interior angles, so 3x = 105° and x = 35°. Since BC ∥ DE, angle D also equals 105° by the same alternate-interior relationship. Only then does the triangle property apply, to triangle DCE: 105° + 24° + y = 180°, giving y = 51°. The habit worth carrying out of this problem: identify every parallel pair in the figure first and extract everything they immediately give for free, before touching any triangle's angle sum at all — trying to apply the triangle property too early, before the parallel lines have supplied angle D, leaves the equation with two unknowns instead of one.

Isosceles Triangles: What Actually Follows From Equal Sides

If it's given that AB = AC in a particular triangle, the base angles opposite those two equal sides are themselves equal — ∠ABC = ∠ACB — but nothing at all forces the third, remaining angle, ∠BAC, to match either of them. Given ∠BAC = 36° specifically, the other two base angles share what's left of the 180° total — 144° — equally between them, 72° each. That single fact, applied where the triangle's sides are extended, cascades outward: an exterior angle at B, ∠ABD, equals ∠BAC + ∠ACB = 36° + 72° = 108°, and inside a smaller triangle formed with a cevian meeting side BD at some point, the angle sum property applies once more to find whatever remains unknown. The general pattern behind all of these isosceles problems is the same: the equal-sides condition supplies one equation (two angles are equal), the angle sum supplies a second (all three add to 180°), and between just those two facts every angle in the triangle becomes solvable, provided at least one angle's actual value is given somewhere in the figure.

Working Entirely Backward From an Exterior Angle

Some problems give only an exterior angle and a ratio between the two interior angles it equals the sum of. If ∠BCD = 125° and ∠A : ∠B = 2:3, then ∠A + ∠B = 125° directly (exterior angle theorem), and splitting 125° across the ratio's 5 total parts gives ∠A = (2/5)×125° = 50° and ∠B = (3/5)×125° = 75° — the ratio only needs to be applied once the exterior angle theorem has already converted the picture into a single sum to divide.

Full Circle Back to Parallel Lines

The angle sum and exterior angle theorems close the loop opened in Exercise 4.3 — the angle sum property is itself proved using a line drawn through one vertex parallel to the opposite side, then reading off alternate interior angles exactly as that exercise practiced. For a refresher on the underlying angle-pair vocabulary this chapter depends on throughout, Exercise 4.2 covers linear pairs and vertically opposite angles in more depth. These same two theorems reappear directly in the Triangles chapter, where they underpin the congruence and similarity criteria studied throughout that chapter.