Geometry — Mathslive.ie

Section 1 of 5

Angle Facts

Most ‘find the angle’ questions come down to a few facts. Mark every angle you can before you solve.

The facts

Straight line: angles add to $180^{\circ}$. Vertically opposite angles are equal. Parallel lines: corresponding (F) equal, alternate (Z) equal, co-interior (C) add to $180^{\circ}$.

Worked example — angles on a line

Three angles on a straight line are $x$, $2x$ and $60^{\circ}$. Find $x$.

$x + 2x + 60 = 180$

$3x = 120$

$x = 40^{\circ}$

Worked example — three angles on a line

Angles $x$, $2x$ and $x$ sit on a straight line. Find $x$.

$x + 2x + x = 180 \Rightarrow 4x = 180$

$x = 45^{\circ}$

Worked example — parallel lines

A transversal crosses two parallel lines and one angle is $70^{\circ}$. Find the corresponding angle $A$ and the co-interior angle.

Corresponding angles are equal, so $A = 70^{\circ}$

Co-interior adds to $180$: $180 - 70$

$A = 70^{\circ}$; co-interior $= 110^{\circ}$

Section 2 of 5

Triangles

Triangle facts

Angles add to $180^{\circ}$. Exterior angle $=$ sum of the two interior opposite. Isosceles: two equal sides $\Rightarrow$ equal base angles. Equilateral: all $60^{\circ}$. Pythagoras: $\text{hyp}^2=\text{opp}^2+\text{adj}^2$. Area $=\tfrac12 bh$.

Worked example — angle sum

A right-angled triangle has angles $90^{\circ}$, $x$ and $2x$. Find $x$.

$90 + x + 2x = 180 \Rightarrow 3x = 90$

$x = 30^{\circ}$

Worked example — an isosceles triangle

An isosceles triangle has a top angle of $70^{\circ}$ and two equal base angles $A$. Find $A$.

$2A + 70 = 180 \Rightarrow 2A = 110$

$A = 55^{\circ}$

Worked example — the exterior angle

An exterior angle of a triangle is $5x+40$; the two interior opposite angles are $60^{\circ}$ and $3x$. Find $x$.

Exterior $=$ sum of interior opposite: $5x+40 = 3x+60$

$2x = 20$

$x = 10^{\circ}$

Worked example — the triangle inequality

A triangle has sides $3$, $5$ and $a$. Find the possible values of $a$.

$a < 3+5 \Rightarrow a < 8$

$5 < a+3 \Rightarrow a > 2$

$2 < a < 8$

Section 3 of 5

Congruent Triangles

Two triangles are congruent when equal in every way. Prove it with one of four tests, then matching sides and angles are equal.

The four tests

SSS three sides · SAS two sides + the angle between · ASA two angles + the side between · RHS right angle, hypotenuse and one side.

Worked example — pick the test

Two triangles have three pairs of equal sides. Which test proves them congruent?

SSS

Section 4 of 5

Parallelograms

Parallelogram facts

Opposite sides and opposite angles are equal. Co-interior angles add to $180^{\circ}$. Area $=$ base $\times$ height. The diagonals bisect each other, and each diagonal bisects the area.

Worked example — a missing angle

One angle of a parallelogram is $70^{\circ}$. Find the angle next to it.

Co-interior: $180 - 70$

$110^{\circ}$

Worked example — a side produced

$abcd$ is a parallelogram with $[dc]$ produced to $e$, and $|\angle bce| = 36^{\circ}$. Find $|\angle abc|$.

$|\angle bcd| = 180 - 36 = 144^{\circ}$ (straight line)

Co-interior with $\angle abc$: $180 - 144$

$|\angle abc| = 36^{\circ}$

Section 5 of 5

The Circle

Learn the parts — centre, radius, diameter, chord, arc, sector, tangent — then the theorems.

Circle theorems

Angle at the centre $=2\times$ angle at the circumference on the same arc. Angles on the same arc are equal. Angle in a semicircle is $90^{\circ}$. Cyclic quadrilateral: opposite angles add to $180^{\circ}$. Tangent $\perp$ radius.

Worked example — central angle

The angle at the centre standing on an arc is $120^{\circ}$. Find the angle at the circumference on the same arc.

Centre $=2\times$ circumference: $\tfrac{120}{2}$

$60^{\circ}$

Worked example — angles on the same arc

Two angles at the circumference stand on the same arc. One is $50^{\circ}$. Find the other, $A$.

$A = 50^{\circ}$ (angles on the same arc are equal)

Worked example — five points on a circle

$a,b,c,d,e$ lie on a circle and $|\angle bde| = 108^{\circ}$. Find $|\angle bae|$ and $|\angle bce|$, with a reason.

$abde$ is a cyclic quadrilateral: $|\angle bae| = 180 - 108 = 72^{\circ}$

$\angle bce$ stands on the same arc $be$ as $\angle bde$, so it is equal

$|\angle bae| = 72^{\circ},\quad |\angle bce| = 108^{\circ}$

Worked example — cyclic quadrilateral

A cyclic quadrilateral has angles $3A$ and $60^{\circ}$ opposite, and $2B$ and $80^{\circ}$ opposite. Find $A$ and $B$.

$3A + 60 = 180 \Rightarrow A = 40^{\circ}$

$2B + 80 = 180 \Rightarrow 2B = 100$

$A = 40^{\circ},\quad B = 50^{\circ}$

Now you try

You try

$A,B,C,D$ lie on a circle. $|\angle ABD|=37^{\circ}$ and $|\angle ADB|=53^{\circ}$. Explain why $[BD]$ is a diameter, and if $|\angle BDC|=46^{\circ}$ find $|\angle CBD|$.

Pen and paper out — try it before you reveal.

$|\angle BAD| = 180 - 37 - 53 = 90^{\circ}$

A $90^{\circ}$ angle stands in a semicircle, so $[BD]$ is a diameter

In $\triangle BCD$, $|\angle BCD|=90^{\circ}$, so $|\angle CBD| = 90 - 46$

$[BD]$ is a diameter; $|\angle CBD| = 44^{\circ}$

You try

$P,Q,R,S$ lie on a circle, centre $O$, with $|\angle PRS|=32^{\circ}$. Find $|\angle SOP|$ and $|\angle SQP|$.

Pen and paper out — try it before you reveal.

Centre $=2\times$ circumference on arc $PS$: $|\angle SOP| = 2(32)$

$\angle SQP$ is on the same arc $PS$, so it equals $\angle PRS$

$|\angle SOP| = 64^{\circ},\quad |\angle SQP| = 32^{\circ}$

You try

$a,d,b,c$ lie on a circle, centre $o$. $|\angle acb|=50^{\circ}$ and $|ad|=|db|$. Find $|\angle aob|$, $|\angle adb|$ and $|\angle oad|$.

Pen and paper out — try it before you reveal.

Centre $=2\times$ circumference: $|\angle aob| = 2(50) = 100^{\circ}$

$abcd$ is a cyclic quadrilateral: $|\angle adb| = 180 - 50 = 130^{\circ}$

$\triangle oad$ is isosceles (two radii): $|\angle oad| = 65^{\circ}$

$|\angle aob|=100^{\circ},\ |\angle adb|=130^{\circ},\ |\angle oad|=65^{\circ}$

You try

A chord $[lk]$ of length $30$ sits in a circle of radius $17$. The radius from the centre meets the chord at right angles at $m$. Find $|om|$.

Pen and paper out — try it before you reveal.

The perpendicular from the centre bisects the chord, so each half is $15$

$|om|^2 + 15^2 = 17^2 \Rightarrow |om|^2 = 289 - 225 = 64$

$|om| = 8$

More examples

You try

A triangle has interior angles $50^{\circ}$ and $60^{\circ}$ at two vertices. Find the exterior angle at the third.

Exterior = sum of the two interior opposite angles.

$50 + 60$

$110^{\circ}$

That’s Geometry.

Angle facts, triangles, congruence, parallelograms and every circle theorem — worked through with a diagram on each example.