bos1234 said:
hmm i typed the quesstion but its not coming
What reason do i give to say that < ROT = < SOT
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this is the real question
< POR = < POS
PRove PQ perp. RS and bisects arc RQS
Ok since
< POR + < ROT = 180* (angle sum of a straight is 180*)
Similarly < POS + < SOT = 180*
But < POR = < POS (given) so < ROT = < SOT
There's no real reasons that need to be given, its just a sylogism (sp)
Like saying A > B ..... B > C ... it logically follows that A > C
Ok consider Triangles ROT and SOT
OT is common
< ROT = < SOT (previous proof)
RO = SO (equal radii)
congurent through two sides and inc. angle being respectively equal.
therefore <ROT = <STO (corresponding angles = on cong. tri)
but
<ROT + <STO = 180* (angle sum of straight)
If 2
<ROT = 180*
<ROT = 90*
PQ perpendicular to RS
Next part:
< ROT = < SOT
arcRQ = arcRS (equal arcs subtend equal angles at the centre)
therefore it bisects RQS
Hope that helps :wave: