Question

Using ruler and compass only, construct a $\Delta$ABC such that $BC=5$cm and AB$=6.5$ and $\angle$ABC $={120}^o$. Construct a cyclic quadrilateral ABCD, such that D is equidistant from AB and BC.

Answer 1

Steps:

1. First draw $BC=5cm$.

2. At $B$, draw an angle of ${120}^o$. Along the direction, mark off $BA=6.5cm$ using compasses.

3. Join $AC$. Thus $△ABC$ is completed.

To sketch the circum-circle, we do the following:

1. Construct the perpendicular bisector of $AB$.

2. Construct the perpendicular bisector of $BC$.

3. Mark the point of intersection of both the perpendicular bisectors as $O$.

4. With $O$ as center and $OA$ as radius, draw a circle. This is the circum-circle of the $△ABC$.

Now, in order to find the point $D$ which is equidistant from $AB$ and $BC$, we construct the angle bisector of $\angle ABC$. Any point on it is equidistant from $AB$ and $BC$. So, we find the point at which this angle bisector intersects the circum circle and name it $D$. Thus $ABCD$ is the required cyclic quadrilateral.

These are shown in the figure attached.