Question

• A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.

Answer 1

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.

Answer 2

$\huge \mathfrak \green{answer \: - }$

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.