In a circle with centre O, chord SR = chord SM. Radius OS intersects the chord RM at P. Prove that RP = PM

given: SM and SR are the equal chords of the circle with center O. OS intersect the chord RM at P.

TPT: RP = PM

proof:

in the triangles OMS and ORS,

OM = OR [radius of the circle]

SM = SR [given]

OS is common.

therefore by SSS congruency triangles OMS and ORS are congruent.

by congruent property: ∠OSM = ∠ OSR..........(1)

now in the triangles SPM and SPR,

SM = SR [given]

∠PSM = ∠ PSR  [using (1) ∠OSM is same as ∠PSM and ∠OSR is same as ∠PSR]

SP is common .

therefore by SAS congruency, triangles SPM and SPR are congruent.

hence PM = RP

which is the required result.

hope this helps you