Question

AB & CD are two equal chord of circle with centre o. If AB and CD, on being produced, meet at a point P outside the circle, Prove that: (i) PA = PC (ii) PB = PD

Answer 1

Draw OM$\perp$AB, ON$\perp$CD and join OP.

Since, perpendicular from the centre bisects the chord.

AM = BM = ½ AB and CN = DN = ½ CD

AM = BM = CN = DN --------- (i)

In $\Delta$OMP and $\Delta$ONP,

1. OM = ON [Equal chords are equidistant from the centre]

2.$\angle$OMP =$\angle$ONP [Each ${90}^0$]

3. OP =OP (common)

$\Delta$OMP $\cong$$\Delta$ONP [by R.H.S]

So, MP = NP (by C.P.C.T) -------- (ii)

1. To prove: PA = PB

MP = NP --------- (ii)

AM = CN -------- (i)

AM + MP = CN + NP

PA = PC (hence, proved)

2. MP = NP ---------- (ii)

BM = DN ---------- (i)

MP – BM = NP – DN

PB = PD (Hence, proved)