8.03a–8.03d, 8.04a, 8.04b - Proof with angles and shapes

8.03a–8.03d, 8.04a, 8.04b - Proof with angles and shapes

A measured diagram can suggest a result, but a proof must establish it for every figure satisfying the given conditions. Build formal chains from angle facts, triangle properties and quadrilateral properties, keeping each reason beside the statement it supports.

Proofs from angle relationships

A diagram of a rhombus makes its diagonals look perpendicular. Would measuring one drawing prove that every rhombus has perpendicular diagonals? Keep that question in mind: the goal is a conclusion forced by the given properties, without a measurement.

A proof starts from given facts, definitions and previously established results. Each step must follow from them. A numerical example can check a proof for a slip, but cannot replace the general argument. In this Higher lesson we will make the links explicit, including facts which a shorter calculation might leave implied.

[DIAGRAM: asset_name: Angle relationships for proof; asset_slug: angle-proof-map; caption: The parallel arrows and the straight lines supply the assumptions for the angle equalities.; description: In the left panel two parallel horizontal lines are crossed at U and V by a rising transversal. The four regions at U are a b c d anticlockwise from the rightward ray, and the four at V are e f g h in the same order. The right panel shows a convex pentagon with one interior angle i and its adjacent exterior angle e, to support the straight-angle relationship at a vertex.]
The parallel arrows and the straight lines supply the assumptions for the angle equalities.

First, even vertically opposite equality can be derived rather than assumed. At U, a+b=180∘a+b=180^\circ and b+c=180∘b+c=180^\circ, because each pair fills a straight angle. Subtracting bb gives a=ca=c. The full turn then gives

d=360∘−(a+b+c)=360∘−(180∘+c)=180∘−c=bd=360^\circ-(a+b+c)=360^\circ-(180^\circ+c)=180^\circ-c=b

The last equality uses b+c=180∘b+c=180^\circ. Thus both opposite pairs are equal, using straight-line and point-angle facts. The lines must genuinely be straight through U; four arbitrary rays do not guarantee opposite equal angles.

Worked example 1: deriving an alternate-angle result

Question. In the parallel-line diagram, prove c=ec=e, using corresponding and vertically opposite angles.

Plan. Link both target angles to the same angle aa.

c=avertically opposite angles at Ua=ecorresponding angles on parallel linesc=eboth equal a\begin{aligned} c&=a && \text{vertically opposite angles at U} \\ a&=e && \text{corresponding angles on parallel lines} \\ c&=e && \text{both equal }a \end{aligned}

Answer. The alternate interior angles cc and ee are equal.

Check. The proof uses the marked parallel relationship in its second line. Without it the two local intersections need not have the same angles, so the general conclusion would fail.

Change only the target pair, from the alternate pair c,ec,e to the co-interior pair c,fc,f. The conclusion becomes a sum rather than an equality.

Worked example 2: deriving a co-interior result

Question. In the same diagram, prove c+f=180∘c+f=180^\circ.

Plan. Use the established equality c=ec=e, then replace ee in the straight angle at V.

c=ealternate angles, proved abovee+f=180∘straight line at Vc+f=180∘substitute c for e\begin{aligned} c&=e && \text{alternate angles, proved above} \\ e+f&=180^\circ && \text{straight line at V} \\ c+f&=180^\circ && \text{substitute }c\text{ for }e \end{aligned}

Answer. The co-interior angles cc and ff are supplementary.

Check. A numerical instance, such as c=61∘c=61^\circ and f=119∘f=119^\circ, satisfies the result. More importantly, no numerical value was needed in the proof, so it applies to every valid configuration.

Polygon angles provide a second kind of formal chain. For a convex nn-sided polygon, each interior angle iki_k and its adjacent exterior turn eke_k total 180∘180^\circ. Adding all nn equalities gives

interior sum+exterior sum=n×180∘\text{interior sum}+\text{exterior sum}=n\times180^\circ

Triangulation gives interior sum (n−2)180∘(n-2)180^\circ, hence

exterior sum=n180∘−(n−2)180∘=360∘\text{exterior sum}=n180^\circ-(n-2)180^\circ=360^\circ

This agrees with one full turn at a point after walking around the boundary. The proof needs integer n≥3n\ge3 and a non-crossing convex boundary for these positive exterior turns. A self-crossing path or unsigned turns around a concave polygon require different care.

Watch out

“The angles look equal” supplies no reason. Replace it with a named relationship and its condition, such as “corresponding angles on the given parallel lines”. Do not use the result you are trying to prove as an unproved starting fact.

Apply the polygon argument without assuming that the five angles are equal.

Proofs using triangle properties

An extra line can expose two triangles whose matching properties force the conclusion. State what is given about that line, then prove its other properties. “Midpoint”, “perpendicular” and “angle bisector” are different conditions; one must not silently stand in for another.

[DIAGRAM: asset_name: Two routes through an isosceles triangle; asset_slug: isosceles-proof-routes; caption: The first proof starts with a midpoint, while the second starts with an angle bisector.; description: Two panels show non-degenerate isosceles triangle ABC with AB equal to AC and D inside segment BC. The left marks BD equal to DC, with AD drawn but no assumed right angle. The right marks angle BAD equal to angle DAC, with AD drawn but no assumed midpoint or right angle. The triangles are labelled to support the stated proofs.]
The first proof starts with a midpoint, while the second starts with an angle bisector.

An angle bisector divides an angle into two equal angles. A segment from a triangle's vertex to the opposite side's midpoint is a median. A perpendicular from a vertex to the line containing the opposite side is an altitude. In a general triangle these can be different lines. For the apex of an isosceles triangle, we can prove that they coincide.

Worked example 3: start with the midpoint

Question. In non-degenerate triangle ABC, AB=ACAB=AC. D is the midpoint of BC. Prove that AD bisects angle BAC and is perpendicular to BC.

Plan. Prove triangles ABD and ACD congruent by SSS, then use corresponding angles and the straight line at D.

AB=ACgivenBD=CDD is the midpointAD=ADcommon side\begin{aligned} AB&=AC && \text{given} \\ BD&=CD && \text{D is the midpoint} \\ AD&=AD && \text{common side} \end{aligned}

Therefore triangles ABD and ACD are congruent by SSS. Corresponding angles give ∠BAD=∠DAC\angle BAD=\angle DAC, so AD bisects angle BAC. Also ∠BDA=∠ADC\angle BDA=\angle ADC. Since B, D and C lie on one straight line, these equal angles sum to 180∘180^\circ; each is 90∘90^\circ.

Answer. AD is both the angle bisector at A and perpendicular to BC.

Check. The proof did not assume the right-angle square it was trying to establish. Non-degeneracy and the midpoint condition put D inside BC and make the two smaller triangles genuine triangles.

Change only the starting condition about AD: give the angle bisector instead of the midpoint. That changes which congruence test is available first.

Continue this lesson.

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