Question

Let $ABC$ be a right triangle in which $AB=6\mathrm{cm}$, $BC=8\mathrm{cm}$ and $\angle B=90°$. $BD$ is the perpendicular from $B$ on $AC$. The circle through $B$, $C$ and $D$ is drawn. Construct the tangents from $A$ to this circle.

Answer 1

Step 1: Construct the triangle

Given, $ABC$ is a triangle with $AB=6\mathrm{cm}$, $BC=8\mathrm{cm}$ and $\angle B=90°$.

1. Draw line $BC=8\mathrm{cm}$.
   
2. Take $B$ as the center and draw an arc of radius $6\mathrm{cm}$.
   
3. Join $AC$.
   

Step 2: Draw the perpendicular

1. Draw an arc centered at $B$ cutting through $AC$ twice.
   
2. Draw arcs of the same length at the newly formed points of intersection such that they meet. Name this point $M$
   
3. Join $BM$. Mark $D$ on $AC$ where $BM$ intersects.
   

Step 3: Draw the circle through $BDC$

1. Draw perpendicular bisector of $BC$ and name the point $E$.
   
2. With $E$ as the center, draw a circle through $B$, $D$ and $C$.
   

Step 4: Draw the tangent

1. Join $AE$.
   
2. Draw perpendicular bisector of $AE$ and name the midpoint as $N$.
   
3. Draw a circle centered at $N$ passing through $A$and $E$. Name the points where it intersects with the circle through $B$, $D$ and $C$ as $Q$ (the other point of intersection will be $B$).
   
4. Join $AQ$.
   

$AQ$ is the required tangent.

Thus, the required tangent is constructed.