Urgent, pls solve it!! I've been out for 1 month for holidays and I forgot everything I did earlier and I'm still trying to patch up but have less time. Pls solve it.
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Answered by
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Hello Mate!
Given : OD is perpendicular from center on chord AB and BC is diameter.
To prove : AC || OD and AC = 2OD.
Proof : Since OD is perpendicular from center on chord so it must also be a line bisector By Theorum.
Hence, AD = BD. Therefore D is mid point on AB.
Similarly, OB and OC are radius.
So, OB = OC
Therefore O is midpoint on diameter BC.
Hence by mid point theorum we get that,
=> OD || AC and OD = ½ AC
=> OD || AC and AC = 2OD
Hence proved ( Q.E.D )
Have great future ahead!
Given : OD is perpendicular from center on chord AB and BC is diameter.
To prove : AC || OD and AC = 2OD.
Proof : Since OD is perpendicular from center on chord so it must also be a line bisector By Theorum.
Hence, AD = BD. Therefore D is mid point on AB.
Similarly, OB and OC are radius.
So, OB = OC
Therefore O is midpoint on diameter BC.
Hence by mid point theorum we get that,
=> OD || AC and OD = ½ AC
=> OD || AC and AC = 2OD
Hence proved ( Q.E.D )
Have great future ahead!
Answered by
0
Step-by-step explanation:
Similarly, OB and OC are radius.
So, OB = OC
Therefore O is midpoint on diameter BC.
Hence by mid point theorum we get that,
=> OD || AC and OD = ½ AC
=> OD || AC and AC = 2OD
Hence proved ( Q.E.D )
Have great future ahead!
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