MathsCBSE Class 10 · Circles

Proofs on Quadrilaterals and Triangles with Tangents

Updated 14 October 2026 · Fact-checked

These are proofs that use the two tangent theorems, tangent perpendicular to radius and equal tangents from an external point, to prove results about shapes around a circle. The board asks you to prove that opposite sides of a circumscribed quadrilateral have equal sums, that tangents at the ends of a diameter are parallel, and related angle results.

Each corner sends two equal tangents (same colour); add the pairs and opposite sides match, so AB + CD = AD + BC.

Illustration drawn for QuestionBuddy.

Understand Proofs on Quadrilaterals and Triangles with Tangents

Every proof in this topic rests on just two facts. Theorem 10.1: the tangent at a point is perpendicular to the radius there. Theorem 10.2: the two tangents from an external point are equal. The skill is not new theorems. It is seeing where these two facts hide inside a figure, then combining them.

Quadrilateral around a circle. If a circle touches all four sides of $A B C D$ , each vertex is an external point sending two tangents. So you get four pairs of equal tangents. Adding them groups the pieces into opposite sides, giving $A B + C D = A D + B C$ . This only holds when the circle touches all four sides. From it, a circumscribed parallelogram must be a rhombus and a circumscribed rectangle must be a square.

Angles at the centre. Joining the centre $O$ to a vertex and to the two contact points makes two right triangles that are congruent by RHS. So $O A$ bisects the angle at the centre between the two radii. Using this at all four vertices, the eight angles at $O$ pair up, and opposite sides subtend supplementary angles at the centre: $∠ A O B + ∠ C O D = 18 0^{\circ}$ .

Parallel tangents. Tangents at the ends of a diameter are both perpendicular to the same line $A B$ , so they are parallel. If a third tangent cuts them at $P$ and $Q$ , then $O P$ and $O Q$ bisect two angles that together make a straight angle, so $∠ P O Q = 9 0^{\circ}$ .

Triangles around a circle. In a triangle with an incircle, equal tangents from each vertex let you swap lengths. For example, an isosceles triangle's incircle touches the base at its midpoint. A circle touching one side and the other two sides produced gives a tangent equal to half the perimeter.

In every proof, write the reason for each equal pair or right angle. The examiner marks the reasons, not just the conclusion.

Proofs on Quadrilaterals and Triangles with Tangents: formulas to remember

Quadrilateral circumscribing a circle: $A B + C D = A D + B C$; Opposite sides have equal sums. Needs the circle to touch all four sides.
Circumscribed parallelogram: $A B = B C = C D = D A$; A parallelogram around a circle is a rhombus; a rectangle around a circle is a square.
Opposite sides subtend supplementary angles at centre: $∠ A O B + ∠ C O D = 18 0^{\circ}, ∠ B O C + ∠ A O D = 18 0^{\circ}$; O is the centre of the circle inscribed in ABCD.
Tangents at ends of a diameter: $l ∥ m$; Both are perpendicular to diameter AB.
Third tangent between parallel tangents: $∠ P O Q = 9 0^{\circ}, P Q = A P + B Q$; P and Q are where the third tangent meets the parallel tangents at A and B.
Angle between tangents and chord-radius angle: $∠ A P B = 2∠ O A B$; PA, PB tangents from P; derive it in a proof, do not just quote it.
Circle touching one side and two sides produced: $A Q = A R = \frac{1}{2} (A B + B C + C A)$; Circle touches BC at P, AB produced at Q, AC produced at R.

How to solve Proofs on Quadrilaterals and Triangles with Tangents questions

Use this routine for any proof or length question built on tangents to a shape.

1Write Given, To prove and, if needed, Construction (usually joining the centre to vertices and contact points).
2Mark every point of contact and write the right angle there, with reason: tangent is perpendicular to radius.
3From each external point (each vertex), list the pair of equal tangents, with reason: tangents from an external point are equal.
4If angles are involved, use RHS congruence to show the line from centre to vertex bisects the angle at the centre.
5Add or subtract the equal pairs so the pieces regroup into the sides or angles you need.
6Use a known total (straight angle $18 0^{\circ}$ , angles round a point $36 0^{\circ}$ , or a given property like opposite sides equal) to finish.
7State the conclusion in the words of the question.

Proofs on Quadrilaterals and Triangles with Tangents in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Quadrilateral circumscribing a circle

Suppose a town park is a quadrilateral ABCD with a circular pond that touches all four boundary fences. The planners write the fence lengths as $A B = (3 x + 1)$ m, $B C = (2 x + 6)$ m, $C D = (x + 9)$ m and $D A = (2 x + 4)$ m. A student tries to find x from the rule for a quadrilateral around a circle and is puzzled.

Show why the rule $A B + C D = B C + D A$ alone cannot fix x here. The total fence length is 60 m: find x, the four sides, and the cost of fencing at ₹150 per metre.

Try it first, then show the solutionSolution

$A B + C D = (3 x + 1) + (x + 9) = 4 x + 10$ and $B C + D A = (2 x + 6) + (2 x + 4) = 4 x + 10$ .
The two sums are equal for every value of x, so the rule gives no equation for x; we need more information.
Perimeter: $(3 x + 1) + (2 x + 6) + (x + 9) + (2 x + 4) = 8 x + 20 = 60$ , so $x = 5$ .
Sides: $A B = 16$ m, $B C = 16$ m, $C D = 14$ m, $D A = 14$ m (check: $16 + 14 = 16 + 14$ ).
Cost of fencing $= 60 \times 150 =$ ₹9000.

Answer: The rule holds for every x, so it cannot find x. With a 60 m perimeter, $x = 5$ : sides 16 m, 16 m, 14 m, 14 m; fencing costs ₹9000.

Memory hook: A pond touching four fences makes opposite fences add up equally, but that alone may not fix the sizes.

Tricks to remember Proofs on Quadrilaterals and Triangles with Tangents

Short memory aids. Say each one aloud once, then use it on a question the same day.

Opposite, not adjacent
Say 'cross the circle': the sides you add are across the circle from each other, $A B$ with $C D$ , $A D$ with $B C$ .
Use it when: Stops the common error of adding adjacent sides.
Two facts, every proof
'Ninety and equal': at a contact point write $9 0^{\circ}$ ; from a vertex write two equal tangents.
Use it when: Start every proof by listing these before thinking further.
Half of a straight line
Two bisected angles that together make $18 0^{\circ}$ give halves adding to $9 0^{\circ}$ .
Use it when: Proving $∠ P O Q = 9 0^{\circ}$ with parallel tangents.
Square at a right corner
A $9 0^{\circ}$ vertex with a circle inside makes a little square, so tangent length $= r$ .
Use it when: Finding tangent segments when a quadrilateral or triangle has a right angle.

Proofs on Quadrilaterals and Triangles with Tangents: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Circumscribed parallelogram is a rhombus · 3 marks · Short answerBoard level

A parallelogram ABCD has a circle touching all four of its sides. Show that the parallelogram must be a rhombus.

Try it first, then show the solutionSolution

Given: parallelogram $A B C D$ touches a circle at $P$ on $A B$ , $Q$ on $B C$ , $R$ on $C D$ , $S$ on $D A$ . To prove: $A B C D$ is a rhombus.
$A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ (tangents from an external point are equal). (1 mark)
Adding: $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$ , so $A B + C D = A D + B C$ . (1 mark)
In a parallelogram $A B = C D$ and $A D = B C$ , so $2 A B = 2 B C$ , giving $A B = B C$ . (½ mark)
Adjacent sides equal in a parallelogram means all sides equal, so $A B C D$ is a rhombus. (½ mark)

Answer: $A B = B C = C D = D A$ , so the parallelogram is a rhombus.

Example 2: Incircle of an isosceles triangle · 3 marks · Short answerBoard level

In $△ A B C$ , $A B = A C$ . A circle inscribed in the triangle touches $B C$ at $D$ , $C A$ at $E$ and $A B$ at $F$ . Prove that $D$ is the midpoint of $B C$ .

Try it first, then show the solutionSolution

$A F = A E$ , $B F = B D$ , $C D = C E$ (tangents from an external point are equal). (1 mark)
Given $A B = A C$ , so $A F + B F = A E + C E$ . (½ mark)
Since $A F = A E$ , subtract to get $B F = C E$ . (½ mark)
So $B D = B F = C E = C D$ . (½ mark)
Hence $B D = C D$ and $D$ is the midpoint of $B C$ . (½ mark)

Answer: $B D = C D$ , so the incircle touches the base at its midpoint.

Example 3: Third tangent between parallel tangents · 3 marks · Short answerBoard level

$A B$ is a diameter of a circle with centre $O$ . Lines $l$ and $m$ are the tangents at $A$ and $B$ . A third tangent touches the circle at $C$ and meets $l$ at $P$ and $m$ at $Q$ , with $P$ and $Q$ on the same side of $A B$ . Prove that $∠ P O Q = 9 0^{\circ}$ .

Try it first, then show the solutionSolution

Join $O C$ . In $△ O A P$ and $△ O C P$ : $O A = O C$ (radii), $∠ O A P = ∠ O C P = 9 0^{\circ}$ (tangent ⊥ radius), $O P$ common. (1 mark)
So $△ O A P ≅ △ O C P$ (RHS), giving $∠ A O P = ∠ C O P$ (CPCT). (½ mark)
Similarly $△ O B Q ≅ △ O C Q$ , giving $∠ B O Q = ∠ C O Q$ . (½ mark)
$A O B$ is a straight line, so $∠ A O C + ∠ B O C = 18 0^{\circ}$ , i.e. $2∠ C O P + 2∠ C O Q = 18 0^{\circ}$ . (½ mark)
So $∠ C O P + ∠ C O Q = 9 0^{\circ}$ , that is $∠ P O Q = 9 0^{\circ}$ . (½ mark)

Answer: $∠ P O Q = 9 0^{\circ}$ .

Example 4: Opposite sides subtend supplementary angles at the centre · 5 marks · Long answerHarder board level

A quadrilateral ABCD has a circle with centre O touching all four of its sides. Show that each pair of opposite sides subtends angles at O that add up to $18 0^{\circ}$ , that is, $∠ A O B + ∠ C O D = 18 0^{\circ}$ and $∠ B O C + ∠ A O D = 18 0^{\circ}$ .

Try it first, then show the solutionSolution

Given: $A B C D$ circumscribes a circle with centre $O$ , touching $A B$ , $B C$ , $C D$ , $D A$ at $P$ , $Q$ , $R$ , $S$ . To prove: $∠ A O B + ∠ C O D = 18 0^{\circ}$ and $∠ B O C + ∠ A O D = 18 0^{\circ}$ . (½ mark)
Construction: join $O P$ , $O Q$ , $O R$ , $O S$ , $O A$ , $O B$ , $O C$ , $O D$ . (½ mark)
In $△ O A P$ and $△ O A S$ : $O P = O S$ (radii), $∠ O P A = ∠ O S A = 9 0^{\circ}$ (tangent ⊥ radius), $O A$ common. So they are congruent (RHS). (1 mark)
Hence $∠ A O P = ∠ A O S$ (CPCT). Call each $a$ . (½ mark)
In the same way $∠ B O P = ∠ B O Q = b$ , $∠ C O Q = ∠ C O R = c$ , $∠ D O R = ∠ D O S = d$ . (1 mark)
Angles round $O$ : $2 a + 2 b + 2 c + 2 d = 36 0^{\circ}$ , so $a + b + c + d = 18 0^{\circ}$ . (½ mark)
$∠ A O B = a + b$ and $∠ C O D = c + d$ , so $∠ A O B + ∠ C O D = 18 0^{\circ}$ . (½ mark)
$∠ B O C = b + c$ and $∠ A O D = a + d$ , so $∠ B O C + ∠ A O D = 18 0^{\circ}$ . (½ mark)

Answer: $∠ A O B + ∠ C O D = 18 0^{\circ}$ and $∠ B O C + ∠ A O D = 18 0^{\circ}$ .

Example 5: Case study: a disc fitted in a badge · 4 marks · Case-basedHOTS / competency

Suppose a school badge is a quadrilateral metal plate $A B C D$ with a circular enamel disc of centre $O$ and radius $6$ cm fixed inside so that it touches all four edges, at $P$ on $A B$ , $Q$ on $B C$ , $R$ on $C D$ and $S$ on $D A$ . $A B = 14$ cm, $B C = 12.5$ cm, $C D = 10.5$ cm, and $∠ A = ∠ D = 9 0^{\circ}$ .
(i) Find $A D$ . (1 mark)
(ii) Find $∠ A O D$ and hence $∠ B O C$ . (1 mark)
(iii) Find $B Q$ and $C R$ , and check your value of $A D$ . (2 marks)

Try it first, then show the solutionSolution

(i) $A B + C D = A D + B C$ (circumscribed quadrilateral), so $14 + 10.5 = A D + 12.5$ , $A D = 12$ cm. (1 mark)
(ii) OA and OD bisect the angles at A and D (equal tangents), so in $△ A O D$ , $∠ O A D + ∠ O D A = 4 5^{\circ} + 4 5^{\circ} = 9 0^{\circ}$ and $∠ A O D = 9 0^{\circ}$ . Opposite sides subtend supplementary angles at the centre, so $∠ B O C = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ . (1 mark)
(iii) At $A$ , $A P O S$ has three right angles and $O P = O S = 6$ cm, so it is a square: $A P = A S = 6$ cm. Similarly at $D$ , $D R = D S = 6$ cm. (½ mark)
$B P = 14 - 6 = 8$ cm, so $B Q = 8$ cm (equal tangents). (½ mark)
$C R = 10.5 - 6 = 4.5$ cm, so $C Q = 4.5$ cm, and $B Q + C Q = 12.5$ cm $= B C$ . (½ mark)
$A D = A S + D S = 6 + 6 = 12$ cm, which matches (i). (½ mark)

Answer: (i) $A D = 12$ cm (ii) $∠ A O D = 9 0^{\circ}$ , $∠ B O C = 9 0^{\circ}$ (iii) $B Q = 8$ cm, $C R = 4.5$ cm; $A D = 12$ cm confirmed.

Example 6: Opposite sides of a circumscribed quadrilateral add up equally · 3 marks · Short answerBoard level

A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively. Prove that $A B + C D = A D + B C$ .

Try it first, then show the solutionSolution

Given: a circle touching AB, BC, CD, DA at P, Q, R, S. To prove: $A B + C D = A D + B C$ . (½ mark)
Tangents from an external point are equal: $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ . (1 mark)
Adding: $A P + B P + C R + D R = A S + B Q + C Q + D S$ . (½ mark)
Grouping: $(A P + P B) + (C R + R D) = (A S + S D) + (B Q + QC)$ , so $A B + C D = A D + B C$ . (1 mark)

Answer: $A B + C D = A D + B C$ .

Where students lose marks in Proofs on Quadrilaterals and Triangles with Tangents

The slips examiners see most often, and what to write instead.

Mistake: Adding adjacent sides instead of opposite sides
Students remember 'sums are equal' but not which sides.
Wrong working
$A B + B C = C D + D A$
Right working
$A B + C D = A D + B C$ (opposite sides)
Do this: Write the four equal-tangent pairs and add them; the grouping comes out automatically.
Usually costs: Usually the whole numerical part of the question.
Mistake: Writing right angles without a reason
The right angle looks obvious in the figure.
Wrong working
$∠ O P A = 9 0^{\circ}$ , so the triangles are congruent.
Right working
$∠ O P A = 9 0^{\circ}$ (tangent at a point is perpendicular to the radius through it).
Do this: Put the reason in brackets on the same line every time.
Usually costs: ½ to 1 mark per missing reason in a proof.
Mistake: Claiming congruence without listing the three pairs
Rushing through the RHS step.
Wrong working
$△ O A P ≅ △ O A S$ , so $∠ A O P = ∠ A O S$ .
Right working
$O P = O S$ (radii), $∠ O P A = ∠ O S A = 9 0^{\circ}$ , $O A$ common; so congruent by RHS, then CPCT.
Do this: Always write the hypotenuse, side and right angle with reasons, then name the rule.
Usually costs: 1 mark in a 5-mark proof.
Mistake: Assuming the tangent length from a right-angled vertex is r without justification
It is true, but the examiner wants the square argument.
Wrong working
$A P = 5$ cm since $r = 5$ .
Right working
$A P O S$ has $∠ A = ∠ P = ∠ S = 9 0^{\circ}$ and $O P = O S$ , so it is a square; $A P = r = 5$ cm.
Do this: Give one line showing the square before using it.
Usually costs: ½ mark.

Exam tips

Proofs in this topic appear as 3-mark short answers and inside 5-mark long answers; write Given, To prove, Construction and Proof clearly.
Describe your figure in words if you cannot draw it neatly: name every centre, contact point and line.
In assertion-reason items, check whether the reason is the exact property used, not just a true statement.
In case studies, the 2-mark part often needs equal tangents chained round the shape; label equal pieces with the same letter first.
Always end a proof with the exact statement asked.

Practice questions on Proofs on Quadrilaterals and Triangles with Tangents

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
A quadrilateral $A B C D$ circumscribes a circle. $A B = (2 x + 1)$ cm, $B C = 3 x$ cm, $C D = (2 x + 3)$ cm and $D A = 10$ cm. Find $x$ and the perimeter of $A B C D$ .
Hint
Set the sums of opposite sides equal.
Answer
$A B + C D = B C + D A$ : $4 x + 4 = 3 x + 10$ , so $x = 6$ . Sides $13, 18, 15, 10$ cm; perimeter $= 56$ cm.
Q2 · 3 marks · Short answerHarder board level
$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ , touching it at $A$ and $B$ . Prove that $∠ A P B = 2∠ O A B$ .
Hint
Let $∠ A P B = θ$ . Use isosceles $△ P A B$ and the right angle $∠ O A P$ .
Answer
$P A = P B$ (equal tangents), so $∠ P A B = ∠ P B A = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$ . $∠ O A P = 9 0^{\circ}$ (tangent ⊥ radius), so $∠ O A B = 9 0^{\circ} - ∠ P A B = \frac{θ}{2}$ . Hence $∠ A P B = 2∠ O A B$ .
Q3 · 3 marks · Short answerHarder board level
A circle touches side $B C$ of $△ A B C$ at $P$ , and touches $A B$ produced at $Q$ and $A C$ produced at $R$ . Prove that $A Q = \frac{1}{2} (A B + B C + C A)$ .
Hint
Split $B C$ at $P$ and swap $B P$ and $C P$ for equal tangents.
Answer
$A Q = A R$ , $B Q = B P$ , $C P = C R$ (equal tangents). Perimeter $= A B + B P + P C + C A = A B + B Q + C R + C A = A Q + A R = 2 A Q$ . So $A Q = \frac{1}{2}$ (perimeter).
Q4 · 5 marks · Long answerHarder board level
$A B$ is a diameter of a circle with centre $O$ . Tangents at $A$ and $B$ are drawn. A third tangent touches the circle at $C$ and meets the tangent at $A$ at $P$ and the tangent at $B$ at $Q$ , with $P$ and $Q$ on the same side of $A B$ . (i) Prove the tangents at $A$ and $B$ are parallel. (ii) If $A P = 4$ cm and $B Q = 9$ cm, find the radius.
Hint
$P Q = P C + C Q$ . $A P QB$ is a right trapezium; drop a perpendicular from $P$ to $B Q$ .
Answer
(i) Both tangents are ⊥ to $A B$ (tangent ⊥ radius), so they are parallel (co-interior angles $9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ ). (ii) $P C = A P = 4$ , $C Q = B Q = 9$ , so $P Q = 13$ cm. Foot $N$ of perpendicular from $P$ on $B Q$ : $N Q = 9 - 4 = 5$ , $P N = A B$ . $A B = 1 3^{2} - 5^{2} = 12$ cm, radius $= 6$ cm. Check: $A P \cdot B Q = 36 = r^{2}$ .
Q5 · 1 mark · Assertion-reasonBoard level
Assertion (A): A parallelogram with sides $7$ cm and $5$ cm cannot circumscribe a circle.
Reason (R): In a quadrilateral circumscribing a circle, the sums of opposite sides are equal.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Add the opposite sides of the parallelogram.
Answer
(a). Opposite sums are $7 + 7 = 14$ and $5 + 5 = 10$ , not equal, so no circle can touch all four sides. R is true and is exactly the reason.
Q6 · 3 marks · Short answerHarder board level
$△ A B C$ is right-angled at $B$ , and a circle with centre O and radius r touches AB, BC and CA. Prove that $r = \frac{A B + B C - A C}{2}$ . Hence find r when $A B = 6$ cm, $B C = 8$ cm and $A C = 10$ cm.
Hint
The radii to the touch points on AB and BC, with the right angle at B, form a square of side r.
Answer
Let the circle touch AB, BC, CA at P, Q, R. $O P ⊥ A B$ , $O Q ⊥ B C$ and $∠ B = 9 0^{\circ}$ , with $O P = O Q = r$ , so OPBQ is a square and $B P = B Q = r$ . Then $A R = A P = A B - r$ and $C R = C Q = B C - r$ . So $A C = A R + R C = A B + B C - 2 r$ , giving $r = \frac{A B + B C - A C}{2}$ . For sides 6, 8, 10: $r = \frac{6 + 8 - 10}{2} = 2$ cm.
Q7 · 5 marks · Long answerHarder board level
A circle of radius 4 cm is inscribed in $△ A B C$ and touches BC at D, with $B D = 8$ cm and $D C = 6$ cm. Find AB and AC.
Hint
Let the tangent length from A be x. Write the area of the triangle in two ways: by Heron's formula, and as the sum of three triangles with O as a common vertex.
Answer
Let the circle touch CA at E and AB at F. Equal tangents: $B F = B D = 8$ , $C E = C D = 6$ , $A E = A F = x$ . So $A B = x + 8$ , $A C = x + 6$ , $B C = 14$ , and the semi-perimeter $s = x + 14$ . (1)
Area as three triangles with vertex O: $\frac{1}{2} \times 4 \times (A B + B C + C A) = 4 s = 4 (x + 14)$ . (1)
By Heron's formula: area $= s (s - a) (s - b) (s - c) = (x + 14) (x) (6) (8) = 48 x (x + 14)$ . (1)
Equating and squaring: $16 (x + 14)^{2} = 48 x (x + 14)$ , so $x + 14 = 3 x$ and $x = 7$ . (1)
Hence $A B = 15$ cm and $A C = 13$ cm. Check: $s = 21$ , area $= 21 \times 7 \times 6 \times 8 = 84 = 4 \times 21$ . (1)
Q8 · 4 marks · Case-basedHOTS / competency
Case study: A circular flower bed with centre O and radius 8 m is touched by two straight hedges PA and PB that start at a gate P, touching the bed at A and B. A straight path DE touches the bed at C, with D on PA and E on PB, cutting off a triangular lawn PDE.
(i) Prove that the perimeter of $△ P D E$ is $2 P A$ . (2 marks)
(ii) The perimeter of the lawn is 30 m. Find PA. (1 mark)
(iii) Find the distance OP of the gate from the centre of the bed. (1 mark)
Hint
Split DE at C and use equal tangents from D and from E.
Answer
(i) Equal tangents: $D C = D A$ and $E C = E B$ . Perimeter $= P D + D E + P E = P D + D C + C E + P E = (P D + D A) + (E B + P E) = P A + P B = 2 P A$ , since $P A = P B$ . (2)
(ii) $2 P A = 30$ , so $P A = 15$ m. (1)
(iii) $∠ O A P = 9 0^{\circ}$ , so $O P = 1 5^{2} + 8^{2} = 289 = 17$ m. (1)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.For $A B C D$ circumscribing a circle, which sums are equal?Show answer
$A B + C D = A D + B C$ .
2.A parallelogram circumscribes a circle. What must it be?Show answer
A rhombus.
3.Why are tangents at the ends of a diameter parallel?Show answer
Both are perpendicular to the same diameter.
4.Third tangent meets parallel tangents at $P$ , $Q$ . What is $∠ P O Q$ ?Show answer
$9 0^{\circ}$ .
5.If $∠ A O B = 10 0^{\circ}$ in a circumscribed $A B C D$ , what is $∠ C O D$ ?Show answer
$8 0^{\circ}$ .

Proofs on Quadrilaterals and Triangles with Tangents: frequently asked questions

Do I need to prove AB + CD = AD + BC every time I use it?

In a numerical question you can quote it as a property of a quadrilateral circumscribing a circle. If the question says prove, write the full working with the four pairs of equal tangents.

Does the opposite sides result work for every quadrilateral?

No. It only holds when a circle touches all four sides. If the sums of opposite sides are unequal, no circle can be inscribed.

Which congruence rule is used in these proofs?

Almost always RHS: the right angle comes from tangent perpendicular to radius, the equal sides are radii, and the hypotenuse is common.

Are constructions of tangents in the syllabus?

No. Constructions of tangents are not in the 2026-27 syllabus, so you only need the proofs and their applications.