In a circle with centre O chord SM = chord SM. Radius is intersects the chord RM at P . Prove that RP = PM
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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 ❤️
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 ❤️
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KaranMaheshwari167:
Figure plz
sorry i forgot!
check it out now I have given!!!
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