Question

Let AB and CD be two equal chords of a circle which intersect within the circle centered at O, as shown below.

Then which of the following is/are true?

• A. AP + DP = CP + BP
• B. AP = CP, but BP $\ne$ DP
• C. BP = DP, but AP $\ne$ CP
• D. AP = CP and BP = DP

Answer 1

Answer: (A), (D)

The correct options are
A

AP + DP = CP + BP

D

AP = CP and BP = DP

Draw OL$\perp$AB and OM$\perp$CD

Join OP.

In $△OPL$ and $△OPM$,

OP = OP (common)

$\angle OLP$ = $\angle OMP$ (Each 90${}^{\circ }$)

OL = OM (Equal chords are equidistant from the centre)

$\therefore △OPL\cong △OPM$ (R.H.S. congruence rule)

$⟹$ PL = PM (c.p.c.t.)

Since, perpendicular from the centre bisects the chord, AL = LB = $\frac{1}{2}$ AB and CM = MD = $\frac{1}{2}$ CD

Also, AL = CM ($\because \frac{1}{2}AB=\frac{1}{2}CD$)

Now, AL - PL = CM - PM

$⟹$ AP = CP

Also, AB - AP = CD - CP

$⟹$ BP = DP

Hence, AP = CP and BP = DP