When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
Step-by-step solution
To find: Show AP = CP and PB = PD (corresponding segments are equal)
Idea: Bring in the perpendiculars from the centre: they bisect the chords and are equal for equal chords. Two right triangles sharing the hypotenuse OP then show P is the same distance from the two midpoints.
- Draw OM ⟂ AB and ON ⟂ CD. These bisect the chords: AM = MB = ½AB and CN = ND = ½CD. Since AB = CD, AM = CN. Name the ends so that M lies between A and P and N between C and P.1 mark
- Equal chords are equidistant from the centre, so OM = ON.½ mark
- In right ΔOMP and ΔONP: ∠OMP = ∠ONP = 90°, hypotenuse OP is common, OM = ON. So ΔOMP ≅ ΔONP (RHS) and MP = NP (CPCT).1 mark
- AP = AM + MP = CN + NP = CP.1 mark
- PB = AB − AP = CD − CP = PD. So the segments of one chord equal the corresponding segments of the other.½ mark
- (If P happens to be the midpoint M, then MP = 0 = NP, so P is also the midpoint of CD and all four segments are equal.)
Check: Numerical check: equal chords AB = CD = 10 cm crossing with AP = 7 cm give CP = 7 cm and PB = PD = 3 cm, and AP × PB = CP × PD = 21, as for any two chords through P.
Answer to write in the exam
Given: AB = CD, chords meeting at P; O the centre. To prove: AP = CP, PB = PD.
Draw OM ⟂ AB, ON ⟂ CD ⇒ AM = ½AB, CN = ½CD (perpendicular from the centre bisects the chord) ⇒ AM = CN
OM = ON (equal chords are equidistant from the centre)
In ΔOMP and ΔONP: ∠OMP = ∠ONP = 90°, OP = OP (common), OM = ON
∴ ΔOMP ≅ ΔONP (RHS) ⇒ MP = NP (CPCT)
AP = AM + MP = CN + NP = CP
∴ AP = CP and PB = AB − AP = CD − CP = PD
Common mistakes that cost marks
- Assuming P is the midpoint of both chords. In general it is not; only the corresponding pieces match.
- Matching the wrong ends: AP equals CP, where A and C are the ends on the same side of the midpoints, not necessarily AP = PD.
- Using SAS for ΔOMP and ΔONP without a known included angle; RHS is the right rule.
How this can come in the exam
Two equal chords AB and CD of a circle, each 10 cm long, intersect at P. If AP = 7 cm (A and C being corresponding ends), find CP and PD.
Show answer
Equal intersecting chords are divided into correspondingly equal segments, so CP = AP = 7 cm (1 mark) and PD = PB = 10 − 7 = 3 cm (1 mark).Try one yourself
Equal chords PQ and RS of a circle are 12 cm long and intersect at X, with QX = 4.5 cm (Q and S corresponding ends). Find PX, RX and SX.
Show answer
SX = QX = 4.5 cm; PX = RX = 12 − 4.5 = 7.5 cm.
More questions like this
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