Question

A point O is taken inside ABCD such that BO = DO. Prove that AOC is a straight

line.​

Answer 1

Answer:

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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