Mixed problems and proofs

Equipped with two theorems and their corollaries, you can tackle a wide variety of board-exam problems. This subtopic walks through a curated set, organised by technique. The recurring habits to internalise are: always draw a clean diagram, mark known lengths and right angles, invoke the two theorems by name, and finish with a clear conclusion line.

Technique 1: Equal tangents in two stages

Many quadrilateral problems use the equal-tangents theorem from each of the four vertices, then add or subtract.

Problem. A circle is inscribed in a triangle $ABC$ with sides $a, b, c$ opposite $A, B, C$ . Show that the tangent lengths from $A, B, C$ to the circle are $s - a, s - b, s - c$ respectively, where $s = (a + b + c)/2$ .

Solution. Let the incircle touch $BC, CA, AB$ at $X, Y, Z$ . Let $AY = AZ = x$ , $BZ = BX = y$ , $CX = CY = z$ (equal-tangents from each vertex). Then $a = BC = y + z$ , $b = CA = z + x$ , $c = AB = x + y$ . Adding: $a + b + c = 2(x + y + z) \Rightarrow x + y + z = s$ . So $x = s - (y + z) = s - a$ , similarly $y = s - b$ , $z = s - c$ . $\blacksquare$

This identity is a workhorse for incircle problems.

Technique 2: The kite $OATB$

When two tangents are drawn from an external point, the figure $OATB$ (centre, contact $A$ , external point $T$ , contact $B$ ) is a kite with two right angles. The angle relationship is

$\angle ATB + \angle AOB = 180^\circ.$

Problem. In a circle of centre $O$ , two tangents from $T$ make $\angle ATB = 80^\circ$ . Find the angle $\angle OAB$ (the angle at the contact point in the triangle $OAB$ ).

$\triangle OAB$ is isosceles (since $OA = OB$ ). Compute $\angle AOB = 180^\circ - 80^\circ = 100^\circ$ . The other two angles of $\triangle OAB$ sum to $80^\circ$ and are equal, so each is $40^\circ$ . So $\angle OAB = 40^\circ$ .

Technique 3: Right triangle Pythagoras

Tangent + radius + line-to-centre forms a right triangle. Pythagoras finishes the problem.

Problem. From a point $13$ cm from the centre of a circle, a tangent of length $12$ cm is drawn. Find the radius.

$r^2 + 12^2 = 13^2 \Rightarrow r^2 = 25 \Rightarrow r = 5$ cm.

Technique 4: Quadrilateral circumscribing a circle

Use $AB + CD = BC + DA$ . Solve for the unknown.

Problem. Quadrilateral $ABCD$ circumscribes a circle. $AB = 6, BC = 7, CD = 4$ . Find $DA$ .

$AB + CD = BC + DA \Rightarrow 10 = 7 + DA \Rightarrow DA = 3$ cm.

Technique 5: Parallel tangent and chord

Problem. A tangent at $P$ to a circle is parallel to a chord $AB$ . Show $PA = PB$ .

Since $OP$ is perpendicular to the tangent at $P$ and the tangent is parallel to $AB$ , $OP \perp AB$ . But the perpendicular from the centre to a chord bisects the chord. So if $OP$ produced meets $AB$ at $M$ , then $AM = MB$ . Now $\triangle PAM \cong \triangle PBM$ (right angle at $M$ , common side $PM$ , $AM = BM$ ), so $PA = PB$ . $\blacksquare$

Technique 6: Tangents and concentric circles

Problem. Two concentric circles of radii $r_1 > r_2$ . A chord of the larger circle is tangent to the smaller. Find the chord length.

The perpendicular from the centre to the chord has length $r_2$ (the radius of the smaller circle, since the chord is tangent to it). The half-chord is $\sqrt{r_1^2 - r_2^2}$ . Full chord: $2\sqrt{r_1^2 - r_2^2}$ .

Worked examples (a fresh batch)

Example 1. In a circle of centre $O$ and radius $5$ , two tangents from $T$ touch the circle at $P$ and $Q$ . If $\angle POQ = 130^\circ$ , find $\angle PTQ$ .

$\angle PTQ = 180^\circ - 130^\circ = 50^\circ$ .

Example 2. Two tangents $TA$ and $TB$ are drawn to a circle of centre $O$ from an external point $T$ . Prove that $\angle AOT = \angle BOT$ .

This follows from the congruence $\triangle OAT \cong \triangle OBT$ (RHS). Corresponding angles equal.

Example 3. A quadrilateral $ABCD$ circumscribes a circle. If $\angle A + \angle C = 180^\circ$ , prove that $\angle B + \angle D = 180^\circ$ as well (and conversely).

Sum of angles in a quadrilateral is $360^\circ$ . Given $\angle A + \angle C = 180^\circ$ , the remainder is $\angle B + \angle D = 180^\circ$ . (This is true of any quadrilateral, regardless of inscribed circle.)

Example 4. A chord $AB$ of a circle subtends an angle of $60^\circ$ at the centre. The tangent at $A$ and the tangent at $B$ meet at $T$ . Find $\angle ATB$ .

$\angle AOB = 60^\circ$ , so $\angle ATB = 180^\circ - 60^\circ = 120^\circ$ .

Example 5. In a triangle $ABC$ , the incircle touches $BC$ at $X$ , $CA$ at $Y$ , $AB$ at $Z$ . If $AB = 10, BC = 14, CA = 12$ , find $AZ$ .

$s = (10 + 14 + 12)/2 = 18$ . $AZ = s - a = 18 - 14 = 4$ cm.

Try it yourself

From an external point $P$ , two tangents to a circle have length $24$ . The radius is $7$ . Find $OP$ .
In a circle, a chord $AB$ subtends $80^\circ$ at the centre. Tangents at $A$ and $B$ meet at $T$ . Find $\angle ATB$ .
A quadrilateral $ABCD$ circumscribes a circle with $AB = 5, BC = 8, CD = 7$ . Find $DA$ .
A triangle has sides $5, 12, 13$ . Find $r$ (inradius).
A tangent at $A$ to a circle is parallel to a chord $BC$ . Prove $AB = AC$ .
Two concentric circles have radii $13$ and $5$ . A chord of the larger circle is tangent to the smaller. Find the chord length.
Two tangents from $T$ to a circle make an angle of $60^\circ$ . Show the chord of contact equals the radius.
In a circle of centre $O$ , $T$ is an external point with $OT = 10$ and tangent length $6$ . Find the radius and $\angle ATB$ .
A right triangle with legs $a, b$ has $r = (a + b - c)/2$ for the incircle, where $c$ is the hypotenuse. Show that this is $\text{Area}/s$ .
Prove that the four vertices of a quadrilateral circumscribing a circle, together with the centre, form a configuration where the four angle bisectors at the vertices meet at the centre.
A chord of length $24$ in a circle of radius $13$ is at distance $d$ from the centre. Find $d$ .
Show that the centre of a circle inscribed in a triangle is the intersection of the angle bisectors.

Pitfalls / Insight

(1) The angle relation $\angle AOB + \angle ATB = 180^\circ$ for the kite is a free mark , never forget it.

(2) For circumscribed quadrilaterals, the sum-of-opposite-sides equality applies to any quadrilateral with an inscribed circle, whether or not it is cyclic.

(3) The incircle "touches at $X, Y, Z$ " with $AX = AY = s - a$ (and cyclic) is one of the most asked configurations. Memorise the tangent-length-from-each-vertex formula.

(4) When writing proofs for the board exam, always cite the theorem you are using: "by the tangent-radius theorem", "by RHS congruence", "by equal tangents from $T$ ". Examiners reward this clarity with full marks.