4.7 - Chains of geometrical reasoning

An angle answer becomes convincing when each step has a reason. You will connect facts about lines, triangles and polygons to find missing values. You will also use chord and tangent facts, with circle-angle reasons in the Higher section.

Link a value to its reason

An isosceles triangle has a base angle of $52^\circ$ . Can you find its exterior angle without measuring the drawing, and convince someone your answer must be right? First, practise attaching a reason to a value.

You already know

Angles on a straight line total $180^\circ$ , and angles around a point total $360^\circ$ . Triangle angles total $180^\circ$ . In $\angle ABC$ , the middle letter $B$ names the vertex.

“I subtracted” describes arithmetic. “Angles around a point total $360^\circ$ ” explains why the subtraction works. Appearance alone does not establish equal lengths, parallel lines or right angles.

Geometrical reason

A geometrical reason is a property that justifies a deduction. For example, vertically opposite angles are equal when two straight lines intersect. “They look equal” is not a geometrical reason.

Choose a property whose conditions are given or already deduced. These properties are absent from both the Foundation and Higher formulae sheets: learn them.

Established situation	Useful reason
A straight line or full turn	Angles total $180^\circ$ or $360^\circ$ , respectively.
Two intersecting straight lines	Vertically opposite angles are equal.
A transversal crosses parallel lines	Alternate or corresponding angles are equal; co-interior angles total $180^\circ$ .
An isosceles triangle	Angles opposite the equal sides are equal.
An $n$ -sided polygon	Interior angles total $(n-2)\times180^\circ$ .
Radius and tangent at the contact point	They meet at $90^\circ$ .
Perpendicular from centre to chord	It bisects the chord. Conversely, the perpendicular bisector of a chord passes through the centre.
Two tangents from one external point	Their lengths are equal.

Worked example 1: a full turn

Question. Two angles that complete one turn are $215^\circ$ and $x^\circ$ . Find $x$ , giving a reason.

Plan. Subtract the known angle from the full-turn total.

\begin{aligned} x+215&=360 &&\text{angles around a point}\\ x&=360-215 &&\text{subtract the known angle}\\ x&=145 &&\text{calculate} \end{aligned}

Why this total? The angles complete a full turn rather than a straight line.

Answer. $x=145$ , because angles around a point total $360^\circ$ .

Check. $215^\circ+145^\circ=360^\circ$ .

Change only the known angle to $235^\circ$ . The reason stays the same; the remaining angle gets smaller.

Worked example 2: a changed angle

Question. Two angles that complete one turn are $235^\circ$ and $y^\circ$ . Find $y$ , giving a reason.

Plan. Use the full-turn total again.

\begin{aligned} y+235&=360 &&\text{angles around a point}\\ y&=360-235=125 &&\text{subtract and calculate} \end{aligned}

Why keep the reason? The size changed, but the arrangement still completes a full turn.

Answer. $y=125$ , using angles around a point.

Check. $235+125=360$ .

A correct number with an unrelated reason does not justify a result.

Watch out

The line $x=180-215$ is wrong here: these angles complete a turn, not a straight line. Use $x=360-215=145$ .

Change only the supplied angle again.

Build and check a linked chain

A chain uses one justified result as the starting point for the next. Name each intermediate angle so your reader knows where each number belongs.

[DIAGRAM: asset_slug: m068_foundation_chain; caption: Isosceles triangle ABC, straight extension B–C–D and parallel line E–A–F; description: AB equals AC. B, C and D are collinear, and EF through A is parallel to BC. The figure is schematic; file: foundation_chain.png; recommended_method: matplotlib]

The equal sides $AB$ and $AC$ face the angles at $C$ and $B$ . The line $AC$ also crosses the parallel lines $EF$ and $BC$ . This transversal lets you carry the angle at $C$ up to $\angle FAC$ .

Worked example 3: equal sides, then angle totals

Question. In the figure, $AB=AC$ , $B,C,D$ are collinear, and $EF\parallel BC$ . Given $\angle ABC=52^\circ$ , find $\angle BAC$ , $\angle ACD$ and $\angle FAC$ , giving reasons.

Plan. Find the other base angle, then use the triangle, straight line and parallel lines.

\begin{aligned} \angle ACB&=52^\circ &&\text{equal base angles}\\ \angle BAC&=180^\circ-52^\circ-52^\circ=76^\circ &&\text{triangle angle sum}\\ \angle ACD&=180^\circ-52^\circ=128^\circ &&\text{straight line}\\ \angle FAC&=52^\circ &&\text{alternate angles, }EF\parallel BC \end{aligned}

Why start with $\angle ACB$ ? Equal sides give it immediately; that value unlocks the later steps.

Answer. The requested angles are $76^\circ$ , $128^\circ$ and $52^\circ$ , respectively.

Check. $52+52+76=180$ , and the exterior angle is $52+76=128$ .

There is a useful surprise in that check: a triangle's exterior angle equals the sum of its two remote interior angles. If those are $p^\circ$ and $q^\circ$ , the third is $(180-p-q)^\circ$ . Its exterior angle is $180-(180-p-q)=p+q$ . This follows from angle totals, rather than measuring a picture.

Change only the given base angle to $58^\circ$ . Keep the same diagram conditions.

Worked example 4: carry a changed base angle through

Question. Under the same figure conditions, $\angle ABC=58^\circ$ . Find $\angle BAC$ , $\angle ACD$ and $\angle FAC$ , giving reasons.

Plan. Repeat the equal-angle deduction before using the angle totals.

\begin{aligned} \angle ACB&=58^\circ &&\text{equal base angles}\\ \angle BAC&=180^\circ-2\times58^\circ=64^\circ &&\text{triangle angle sum}\\ \angle ACD&=180^\circ-58^\circ=122^\circ &&\text{straight line}\\ \angle FAC&=58^\circ &&\text{alternate angles, }EF\parallel BC \end{aligned}

Why not use $360^\circ$ at $C$ ? The interior and exterior angles together make a straight angle.

Answer. The angles are $64^\circ$ , $122^\circ$ and $58^\circ$ .

Check. The remote angles give $64^\circ+58^\circ=122^\circ$ .

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Number

Algebra

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Handling data

4.7 - Chains of geometrical reasoning

Link a value to its reason

Build and check a linked chain

Continue this lesson.

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