6.2 - Geometric proof

6.2 - Geometric proof

Build geometric arguments from stated facts, with a reason for every deduction. You will combine angle sums, parallel lines, circle theorems and congruent triangles to prove results for whole families of shapes. The diagram suggests a route; the argument establishes it.

Formal angle arguments

Can a sloping parallelogram have all four vertices on one circle? A carefully drawn example might persuade you either way. A proof will settle every case, using just two angle facts.

You already know

Triangle angles total 180∘180^\circ, angles on a straight line total 180∘180^\circ, and an isosceles triangle has equal base angles. Parallel lines give equal alternate or corresponding angles and supplementary co-interior angles.

A formal argument starts from facts given in the question or already established by a theorem. It names the angle or length it deduces and gives the reason. A measured angle is evidence about one drawing, not a proof that the result always holds. Likewise, writing the desired conclusion at the start does not establish it.

Geometrical proof

A geometrical proof is a connected argument showing that a conclusion follows from stated conditions and valid geometric results. A proof that uses parallel-line angles must identify the parallel lines; a proof using a circle theorem must identify the relevant circle and chord or tangent.

When exact values are absent, name angles with letters. The same equation then covers every permissible shape. The geometric facts used here are not on the formulae sheet: learn them.

Worked example 1: proving the exterior-angle theorem

Question. In triangle ABCABC, side BABA is extended beyond AA to DD. Prove that ∠DAC=∠ABC+∠BCA\angle DAC=\angle ABC+\angle BCA.

Plan. Express the interior angle at AA using the triangle sum, then use the straight line BADBAD.

Let ∠ABC=β\angle ABC=\beta and ∠BCA=γ\angle BCA=\gamma.

∠BAC=180∘−β−γangles in triangle ABC∠DAC=180∘−∠BACB,A,D collinear=180∘−(180∘−β−γ)substitute the whole angle=β+γsimplify\begin{aligned} \angle BAC&=180^\circ-\beta-\gamma && \text{angles in triangle }ABC \\ \angle DAC&=180^\circ-\angle BAC && B,A,D\text{ collinear} \\ &=180^\circ-(180^\circ-\beta-\gamma) && \text{substitute the whole angle} \\ &=\beta+\gamma && \text{simplify} \end{aligned}

Why this step? Both equations contain the same interior angle, so substitution connects the straight-line and triangle facts.

Answer. ∠DAC=∠ABC+∠BCA\angle DAC=\angle ABC+\angle BCA, as required.

Check. If β=48∘\beta=48^\circ and γ=67∘\gamma=67^\circ, the interior angle at AA is 65∘65^\circ and the exterior angle is 115∘115^\circ. This supports the general argument but does not replace it.

Worked example 2 adds just one condition: the triangle is isosceles. That turns a sum of two angles into twice one angle.

Worked example 2: using an equal-side condition

Question. In triangle ABCABC, AB=ACAB=AC. Side BABA is extended beyond AA to DD. Prove that ∠DAC=2∠ABC\angle DAC=2\angle ABC.

Plan. Use equal base angles, then apply the exterior-angle result.

Let ∠ABC=β\angle ABC=\beta.

∠BCA=βAB=AC gives equal base angles∠DAC=∠ABC+∠BCAexterior angle theorem=β+β=2βsubstitute\begin{aligned} \angle BCA&=\beta && AB=AC\text{ gives equal base angles} \\ \angle DAC&=\angle ABC+\angle BCA && \text{exterior angle theorem} \\ &=\beta+\beta=2\beta && \text{substitute} \end{aligned}

Why this step? Equal sides tell you which two angles are equal: each is opposite one of those sides.

Answer. ∠DAC=2∠ABC\angle DAC=2\angle ABC, as required.

Check. With β=52∘\beta=52^\circ, the apex angle is 76∘76^\circ and its exterior angle is 104∘=2(52∘)104^\circ=2(52^\circ).

A reason such as “triangle” is too vague to show what you used. “Angles in triangle ABCABC total 180∘180^\circ” states a usable fact.

Watch out

Wrong: assume ∠DAC=2β\angle DAC=2\beta, then rearrange to obtain equal base angles. That assumes the conclusion. Right: start with the given equal sides, deduce the base angles and finish with the desired exterior angle.

Move the equal-sided vertex to a different position and rebuild the reasoning.

Combining shape conditions

Two familiar properties can force a much stronger conclusion when they apply to the same pair of angles.

A cyclic quadrilateral has opposite angles totalling 180∘180^\circ. A parallelogram has opposite angles equal. If both descriptions hold, the same two angles must be equal and supplementary. Neither condition alone forces right angles; together they do.

Equal and supplementary angles

If α=β\alpha=\beta and α+β=180∘\alpha+\beta=180^\circ, then

2α=180∘,α=β=90∘2\alpha=180^\circ,\qquad\alpha=\beta=90^\circ

This is a deduction from the two stated conditions, not an extra circle theorem.
Not on the formulae sheet: learn it.

This is the answer to the opening puzzle waiting to happen. Name one opposite pair and apply both conditions to it.

Worked example 3: a cyclic parallelogram

Question. ABCDABCD is both a parallelogram and a cyclic quadrilateral. Prove that it is a rectangle.

Plan. Combine equal opposite angles with supplementary opposite angles, then establish all four right angles.

Let ∠DAB=α\angle DAB=\alpha.

∠BCD=αopposite angles of a parallelogramα+α=180∘opposite angles of a cyclic quadrilateralα=90∘solve∠ABC=∠CDA=90∘adjacent parallelogram angles supplementary\begin{aligned} \angle BCD&=\alpha && \text{opposite angles of a parallelogram} \\ \alpha+\alpha&=180^\circ && \text{opposite angles of a cyclic quadrilateral} \\ \alpha&=90^\circ && \text{solve} \\ \angle ABC&=\angle CDA=90^\circ && \text{adjacent parallelogram angles supplementary} \end{aligned}

Why this step? The two shape conditions refer to the same opposite pair, so their conclusions can be combined directly.

Answer. All four angles are right angles, so ABCDABCD is a rectangle.

Check. A non-rectangular parallelogram with angles 70∘,110∘,70∘,110∘70^\circ,110^\circ,70^\circ,110^\circ fails the cyclic condition: its opposite 70∘70^\circ angles sum to 140∘140^\circ.

Worked example 4 strengthens just one condition: the parallelogram has all four sides equal, so it is a rhombus.

Worked example 4: a cyclic rhombus

Question. ABCDABCD is a rhombus whose vertices lie on a circle. Prove that it is a square.

Plan. Use the parallelogram angle argument, then retain the rhombus's equal sides.

Let ∠DAB=α\angle DAB=\alpha.

∠BCD=αa rhombus is a parallelogram2α=180∘cyclic opposite anglesα=90∘solve∠ABC=∠CDA=90∘adjacent angles supplementary\begin{aligned} \angle BCD&=\alpha && \text{a rhombus is a parallelogram} \\ 2\alpha&=180^\circ && \text{cyclic opposite angles} \\ \alpha&=90^\circ && \text{solve} \\ \angle ABC&=\angle CDA=90^\circ && \text{adjacent angles supplementary} \end{aligned}

Why this step? Equal sides alone do not make a square; the cyclic condition supplies the missing right angles.

Answer. The quadrilateral has four equal sides and four right angles, so it is a square.

Check. A rhombus with an angle of 80∘80^\circ would have the opposite angle 80∘80^\circ, contradicting the cyclic sum 180∘180^\circ.

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