Question

Plzz give answer.


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

Answer 1

$\huge\red{ANSWER}$

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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.

$\Huge{\boxed{\mathcal{\red{THANKYOU}}}}$

Answer 2

Answer:

see below

Step-by-step explanation:

Join O with all vertices

now ∆ABO =~ ∆ ADO ( SSS axiom )

hence <AOD = < AOB = x° ( CPCT).........(1)

also ∆COD =~ ∆ BOC ( SSS axiom )

hence <BOC = < COD = y° ( CPCT).........(2)

but <AOD + < AOB + <BOC + < COD = 360° ( complete angle)

means...x° + x° + y° + y° = 360°

2 ( x° + y° ) = 360°

x° + y° = 180°

therefore...< AOD + <COD = 180°

hence....AOC is a straight line

proved