Circle theory, chords, quadrilaterals and circumcenter

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• (a) State the theorem on two intersecting chords of a circle and the parts into which they are divided. Draw a careful figure with hypothesis and thesis, and indicate how to prove it.
• (b) Two chords $AB$ and $CD$ meet at $K$. Join $A$ with $C$ and $B$ with $D$: prove that $\widehat{C\!AB}\cong\widehat{B\!DC}$; then draw the bisectors of those angles and prove they meet on the circle (hint: by contradiction).
• (c) State the necessary and sufficient condition for a quadrilateral to be cyclic and for it to be tangential, with examples.
• (d) Prove that in a right triangle the circumcenter is the midpoint of the hypotenuse.
• (e) Prove that if two chords meet perpendicularly and bisect each other, they are the diagonals of an inscribed square.
• (f) Draw a quadrilateral that is both cyclic and tangential, but not a square.

Solution

(a) Intersecting chords theorem. If two chords $AB$ and $CD$ meet at an interior point $K$, then $\overline{AK}\cdot\overline{KB}=\overline{CK}\cdot\overline{KD}$. Hypothesis: $AB,CD$ chords, $K=AB\cap CD$ interior. Thesis: $AK\cdot KB=CK\cdot KD$. Proof: triangles $AKC$ and $DKB$ are similar since they have vertical angles at $K$ and $\widehat{CAK}\cong\widehat{BDK}$ (inscribed angles on the same arc $BC$); from $\dfrac{AK}{DK}=\dfrac{CK}{BK}$ the thesis follows.

(b) $\widehat{CAB}$ and $\widehat{BDC}$ are inscribed angles both standing on arc $BC$ (on the same side), hence congruent. Their bisectors each cut arc $BC$ at its midpoint; since the angles are equal and stand on the same arc, both bisectors pass through the same midpoint $M$ of arc $BC$: they meet on the circle. (By contradiction, if they met at a different interior point, the two inscribed angles would not subtend the same arc.)

(c) A quadrilateral is cyclic $\iff$ opposite angles are supplementary ($\hat A+\hat C=\hat B+\hat D=180^\circ$). It is tangential $\iff$ the sums of opposite sides are equal ($AB+CD=BC+DA$, Pitot’s theorem). Examples: a rectangle is cyclic; a rhombus is tangential.

(d) Let $ABC$ be right-angled at $C$. The circumcenter is equidistant from the vertices; the hypotenuse $AB$ subtends the right angle at $C$, so $AB$ is a diameter of the circumscribed circle (a $90^\circ$ inscribed angle stands on a semicircle). The centre of diameter $AB$ is its midpoint $M$, equidistant from $A,B,C$: $M$ is the circumcenter. $\;\blacksquare$

(e) Let $AB$ and $CD$ be chords meeting at $K$, perpendicular and bisecting each other ($AK=KB$, $CK=KD$). Since each chord is bisected by $K$, the line through the centre $O$ and $K$ is perpendicular to each chord: hence $O=K$, both are diameters, hence congruent. Perpendicular, congruent, mutually bisecting diagonals $\Rightarrow$ $ACBD$ is an inscribed square. $\;\blacksquare$

(f) A right kite (two pairs of consecutive equal sides, with a pair of opposite right angles) is both cyclic and tangential, but if not equilateral it is not a square. $\boxed{AK\cdot KB=CK\cdot KD;\quad \text{cyclic}\!:\hat A+\hat C=180^\circ;\quad \text{tangential}\!:AB+CD=BC+DA.}$