Question

Let $ABC$ be a right triangle in which $AB=6cm$, $BC=8cm$ and $\angle B=90°$.

$BD$ is the perpendicular from $B$ on $AC$.

The circle through $B,C,D$ is drawn.

Construct the tangents from $A$ to this circle.

Give the justification of the construction

Answer 1

Construct the tangents to the given circle

Following steps are followed to perform the required construction:

1. Draw a line segment $BC=8cm$
2. Draw an a ray from vertex $B$ such that the ray is perpendicular to $BC$
3. With center $B$. draw an arc of radius $6cm$ which cuts the ray at point $A$
4. Join the points $A,C$ to complete $△ABC$
5. Draw the perpendicular bisector of $BC$ to locate its midpoint $E$
6. With $E$ as center and $BE$ as radius draw a circle.
7. This circle will intersect the triangle at point $D$. Join $BD$
8. Join $AE$ and draw the perpendicular bisector to $AE$ to locate its midpoint $M$.
9. With $M$ as center and $AM$ as radius draw a circle.
10. This circle will intersect the previous circle at points $B,Q$
11. Join $AQ$
12. $AB$ and $AQ$ are required tangents.

Justification of the construction:

1. $BE,BQ$ are radii of the first circle, hence tangents from $A$ must be perpendicular to these radii at $B,Q$
2. Angle in a semicircle is always

Hence, we draw a circle with $AE$ as diameter so it will give to semicircles which will give ${90}^{\circ }$ angles.

Thus, the given construction problem is completed.