Question

O is the centre of circle, chords MN and RS are intersecting at P.if OP is the bisector of AngleMPR,prove that MN=RS​

Answer 1

Step-by-step explanation:

first approve 90 degree then prove the triangle similar and by then op is bisector angle MPR

so , MN=RS

Answer 2

O is the centre of circle, chords MN and RS are intersecting at P.if OP is the bisector of AngleMPR, then MN=RS​.

Drop a perpendicular from centre O to the chords MN and RS. Let the foot of perpendiculars be X and Y respectivly.

In ΔPXO and ΔPYO,

• ∠XPO = ∠YPO , angle bisector is OP

• ∠PXO = ∠PYO = 90 degree

• OP = OP , common side.

Therefore by AAS Congruence rule,

• ΔPXO ≅ ΔPYO

• This implies OX = OY , 2 corresponding sides of congruent triangles.

Consider in ΔMXO and ΔRYO

• since , sin( ∠XMO )  = radius/perpendicular = OM/OX

• also sin (∠YRO ) = radius / perpendicular = OR/OY

• Since OR = OM = radius of circle

• and OY = OX , proved earlier.

• Therefore ∠NMO = ∠SRO

Therefore by AAS congruence rule,

• ΔMXO = ΔRYO

Therefore MX = RY

==> If length of perpendicular from the center to two chords are equal, then the length of chords are equal.

We know perpendicular from the centre to the chords of a circle, bisects  the chord.

Therefore, 2MX = MN = 2RY = RS

Thus proved that MN = RS