Question

tell me the answer..​

Answer 1

Answer:

$\sf\mathbb\color{red} {⟹ BD : AC: : 3 : 1 }$

Step-by-step explanation:

Given: ABCD is a parallelogram, where AC and BD are the diagonals meeting at O. AB = BC = AC.

To Prove: BD : AC :: √3 : 1

Proof : In △ABC, AB = BC = CA (given).

= a (say)

Hence ABC is an equilateral triangle . (definition of equilateral triangle)

AC and BD are the diagonals of parallelogram ABCD.

⇒ AC = BD (Diagonals of a parallelogram bisect each other)

or AO = OC.

i.e. BO is the median of the equilateral ABC.

Hence BO = √3/2a

∴ BD = √3a

⇒ BD : AC :: √3 a : a

⇒ BD : AC :: √3 : 1

Answer 2

Answer:

Given: ABCD is a parallelogram, where AC and BD are the diagonals meeting at O. AB = BC = AC.

To Prove: BD : AC :: √3 : 1

Proof : In △ABC, AB = BC = CA (given).

= a (say)

Hence ABC is an equilateral triangle . (definition of equilateral triangle)

AC and BD are the diagonals of parallelogram ABCD.

⇒ AC = BD (Diagonals of a parallelogram bisect each other)

or AO = OC.

i.e. BO is the median of the equilateral ABC.

Hence BO = √3/2a

∴ BD = √3a

⇒ BD : AC :: √3 a : a

⇒ BD : AC :: √3 : 1