Question

Let the line segments $AB$ and $CD$ intersect at $O$ is such a way that $OA=OB=$ $OC$, then $AC=$ BD but $AC$ may not be parallel to $BD$.

• A. True
• B. False

Answer 1

Answer: (A)

It is given that,

$AO=OD$ and $CO=OB$

Here, line segment $AB$ and $CD$ are concurrent.

So,

$\angle AOC=\angle BOD$ [ Vertically opposite angle ]

Now,

In $△AOC$ and $△DOB,$

$\Rightarrow$ $AO=OD$

$\Rightarrow$ $CO=OD$

$\Rightarrow$ $\angle AOC=\angle BOD$

$\therefore$ $△AOC\cong △BOD$ [ By $SAS$ property of congruence ]

$\Rightarrow$ $AC=BD$ [ C.P.C.T ]

Here,

$\angle ACO\ne \angle BDO$ or $\angle OAC\ne \angle OBD$

Hence, there are no alternate angles, unless both triangles are isosceles triangle.

Hence, it is proved that, $AC=BD$ but $AC$ may not be parallel to $BD.$