Four Corners on One Circle
You already know that the perpendicular bisectors of a triangle meet at a single point, and that this point is the same distance from all three vertices — so every triangle has a circumscribed circle passing through its three corners. Quadrilaterals are not so lucky. Pick four points at random and the circle through three of them will usually miss the fourth. A cyclic quadrilateral is the special, well-behaved case where all four vertices do happen to sit on one circle. The circle is called its circumcircle, and the quadrilateral is said to be inscribed in it.
Why care about this one case out of all quadrilaterals? Because the moment the corners of a cyclic quadrilateral are pinned to a circle, the inscribed angle theorem from the previous guide takes over and forces a beautiful rigidity on the angles. Each of the four corner angles is an inscribed angle looking across at the arc on the far side, and inscribed angles are completely controlled by their arcs. That single fact is the engine for everything in this guide.
Opposite Angles Add to 180
Here is the headline property. In a cyclic quadrilateral ABCD, the two pairs of opposite angles are supplementary: m(angle A) + m(angle C) = 180 degrees, and likewise m(angle B) + m(angle D) = 180 degrees. The proof is short and rests entirely on a central angle being twice the inscribed angle that shares its arc. Watch how the whole circle of 360 degrees gets split exactly in half.
- Angle A and angle C sit at opposite corners. Angle A is inscribed in the arc on its side and opens onto the far arc BCD; angle C opens onto the far arc DAB. Together those two far arcs are the whole circle, so their measures add to 360 degrees.
- By the inscribed angle theorem, each angle is half its far arc: m(angle A) is half of arc BCD, and m(angle C) is half of arc DAB.
- Add them: m(angle A) + m(angle C) is half of (arc BCD + arc DAB), which is half of 360 degrees, namely 180 degrees.
The other pair, angle B and angle D, must then also sum to 180 degrees, since all four interior angles of any quadrilateral add to 360 degrees and we have just accounted for half of it. So a cyclic quadrilateral is angle-locked: tell me one angle and I instantly know the one across from it. A rectangle is the friendliest example — its four right angles obviously pair up to 180, which is exactly why every rectangle is cyclic, with the circle centred at the meeting of its diagonals.
Running It Backwards: A Test for Being Cyclic
A theorem and its converse are different claims, as you learned when you first separated a statement from its converse — and here the converse happens to be true and genuinely useful. If a quadrilateral has one pair of opposite angles summing to 180 degrees, then it must be cyclic: a circle passes through all four of its vertices. This gives you a quick certificate. You do not have to find the circle; you just check one angle sum.
The idea behind the converse is a clean little contradiction. Draw the circle through three of the vertices, say A, B, C. If the fourth vertex D were not on that circle, it would lie either inside or outside it; in either case the angle at D would be measurably more or less than the value the supplementary condition demands, contradicting m(angle B) + m(angle D) = 180. So D has nowhere to go but onto the circle. We will not chase every inequality here — the full case analysis is honest but fiddly — yet the shape of the argument is exactly the locus thinking you have already met: the points seeing BC at a fixed angle form an arc, and D is pinned to it.
The Power of a Point: One Number, Many Lines
Now fix a point P and a circle, and draw any line through P that crosses the circle, hitting it at two points X and Y. The power of a point is the remarkable claim that the product |PX| times |PY| is the same number for every such line through P — it does not depend on the line you chose, only on where P sits relative to the circle. Swing the line around to a new direction and the two intersection distances individually change, but their product holds perfectly still.
Where does this constancy come from? From cyclic quadrilaterals, fittingly. Take two chords through P, one meeting the circle at A and C, the other at B and D. The four points A, B, C, D lie on the circle, so ABCD is cyclic and its inscribed angles are tied to arcs. Compare triangle PAB with triangle PDC: they share the angle at P, and the inscribed angles angle A and angle D both subtend the same arc BC, so they are equal. Two equal angles make the triangles similar by AA. Similar triangles have proportional sides, and writing that proportion out and cross-multiplying gives |PA| times |PC| = |PB| times |PD|. The product is the same for both chords — exactly the power of the point.
P inside the circle (two chords crossing at P):
A
\ |PA| . |PC| = |PB| . |PD|
B---P---D (the common value is the POWER of P)
\
C
P outside the circle (two secants from P):
P----A----C |PA| . |PC| = |PB| . |PD|
\
B----D and if one line is a TANGENT at T:
|PT|^2 = |PA| . |PC|Inside, Outside, and the Tangent Limit
The same statement wears three faces depending on where P sits. If P is inside the circle, every line through it is a chord, and you get the intersecting-chords rule: |PA| times |PC| = |PB| times |PD|, two pieces of one chord multiplied equal two pieces of the other. If P is outside, the lines through it that hit the circle are secants, each entering and leaving, and the same product rule holds for the two secants. The cleanest way to keep them straight is to remember they are one theorem, proved by one pair of similar triangles, just drawn in two positions.
The prettiest face appears in the limit. Slide one of the secants from an outside point P until its two intersection points slowly merge into a single point T — the line has become a tangent, touching the circle at exactly T. The product |PX| times |PY| collapses into |PT| times |PT|, that is |PT|^2. So from an external point, the square of the tangent length equals the secant product: |PT|^2 = |PA| times |PC|. This tangent-secant relation is the workhorse version, and it leans quietly on the tangent meeting the radius at a right angle and on the tangent-chord angle equalling the inscribed angle in the alternate segment, both from earlier in this rung.
Putting It to Work, Honestly
These two results turn awkward geometry problems into one-line arithmetic. Suppose a chord of length 10 is cut by another chord into pieces of 2 and 8 on one side; if the second chord is split into x and 6, then the power rule says 2 times 8 = x times 6, so x = 16/6, a little over 2.6 — no angle-chasing required. Or, from a point P outside a circle, if a tangent has length 6 and a secant from P cuts the circle at distances 4 and L, then 6^2 = 4 times L gives L = 9. The power of a point is a quiet bookkeeping rule that the messy picture must obey.
Two honest cautions to close the rung. First, every theorem here needs the points genuinely on the circle, or the lines genuinely through the one chosen point P — move a corner off the circle and the angle-sum fails; let two chords cross at different points and the product rule is simply about a different quantity. Second, the inscribed-angle machinery underneath all of this carries its own fine print from the previous guide: an inscribed angle equals half the central angle only when it subtends the same arc, and equal inscribed angles must stand on the same side of their chord. Honour those conditions and the cyclic quadrilateral and the power of a point will never let you down.