Question

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

Asap plz

Answer 1

Answer:

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

To Proof: RP = PM

Proof:

In the ∆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.

B CPCT: ∠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!

^_^

Happy Valentine's Day ❤️