Q6:

In each of the following, give also the justification of the construction: Let ABC be a right triangle in which $AB = 6$ cm, $BC = 8$ cm and $\angle B = 90^{\circ}$. $BD$ is the perpendicular from $B$ on $AC$. The circle through $B$, $C$, $D$ is drawn. Construct the tangents from $A$ to this circle.

Given: A right-angled triangle $ABC$ where $AB = 6$ cm, $BC = 8$ cm, and $\angle B = 90^{\circ}$. $BD$ is the altitude from $B$ to the hypotenuse $AC$. A circle is drawn passing through points $B$, $C$, and $D$.

To Find: Construct the tangents from point $A$ to the circle passing through $B$, $C$, and $D$, and provide the justification.

Step 1: Construction of Triangle ABC

1. Draw a line segment $BC = 8$ cm.

2. At point $B$, construct an angle of $90^{\circ}$ using a compass and ruler.

3. Cut an arc of $6$ cm on the perpendicular line from $B$ to mark point $A$.

4. Join $AC$. This forms the right-angled triangle $ABC$.

Step 2: Construction of the Circle

1. Draw a perpendicular from $B$ to $AC$ to locate point $D$. [Since $BD \perp AC$, $\angle BDC = 90^{\circ}$].

2. Since $\angle BDC = 90^{\circ}$, the segment $BC$ subtends a right angle at $D$. Therefore, $BC$ is the diameter of the circle passing through $B, D,$ and $C$.

3. Find the midpoint $O$ of $BC$.

4. With $O$ as the center and $OB$ as the radius, draw a circle. This circle passes through $B, D,$ and $C$.

Step 3: Construction of Tangents from A

1. Join $AO$.

2. Find the midpoint $M$ of $AO$ by drawing the perpendicular bisector of $AO$.

3. With $M$ as the center and $MA$ as the radius, draw a circle. This circle intersects the circle with center $O$ at two points. One point is $B$, and let the other point be $E$.

4. Join $AE$. $AB$ and $AE$ are the required tangents.

Justification:

1. $AB$ is already a tangent to the circle because $\angle ABO = 90^{\circ}$ (given in the triangle construction). Since $OB$ is the radius, a line perpendicular to the radius at its endpoint on the circle is a tangent.

2. For the second tangent $AE$: In the circle with center $M$, $\angle AEO = 90^{\circ}$ [Angle in a semicircle].

3. Since $OE$ is a radius of the circle with center $O$, and $AE \perp OE$ at point $E$ on the circle, $AE$ must be a tangent to the circle.

Final Answer: The tangents from point $A$ to the circle are the line segments $AB$ and $AE$, where $B$ is the vertex of the right angle and $E$ is the point of intersection of the auxiliary circle with the circle passing through $B, C, D$.