plz solve question 19
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ok I will try it.......
So,,
HERE'S YOUR ANSWER...
A point O is taken inside a rhombus such that its distance from the vertices B and D are equal. How can you show AOC is a straight line?
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GIVEN: A rhombus ABCD. AC & BD diagonals meet at P.
Since , ABCD is a rhombus
=> AC & BD diagonals are perpendicular bisectors to each other.
Point O is given inside ABCD, such that OD = OB.
TO PROVE: AOC is a straight line.
PROOF: Since O is equidistant from D & B
=> O lies on the perpendicular bisector of segment joining D & B ie segment DB.
And P also lies on the perpendicular bisector of DB.
=> OP is the perpendicular bisector of DB.
But, given that AP is perpendicular bisector of DB.
=> OP coincides with AC ( as a segment can not have 2 distinct perpendicular bisectors)
=> A,O,P,C are collinear.
Hence, AOC is a straight line.
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Point O is inside rhombus ABCD such that it is equidistant from points B and D. We want to show that AOC is a straight line.
BD is a diagonal of rhombus ABCD and point O is equidistant from points B and D.
⇒ Point O lies on the perpendicular bisector of diagonal BD.
One of the properties of a rhombus is that its diagonals are perpendicular bisectors of each other.
⇒ Point O lies on the line joining the points A and B.
⇒AOC is a straight line.
HOPE IT WILL HELP YOU.
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