Let $\mathrm{ABC}$ be a right triangle in which $\mathrm{AB}=6 \mathrm{~cm}, \mathrm{BC}=8 \mathrm{~cm}$ and $\angle \mathrm{B}=90^{\circ} . \mathrm{BD}$ is the perpendicular from $\mathrm{B}$ on $\mathrm{AC}$. The circle through B, C,D is drawn. Construct the tangents from A to this circle.
Consider the following situation. If a circle is drawn through $\mathrm{B}, \mathrm{D}$, and $\mathrm{C}, \mathrm{BC}$ will be its diameter as $\angle \mathrm{BDC}$ is of measure $90^{\circ}$. The centre $\mathrm{E}$ of this circle will be the midpoint of BC.
The required tangents can be constructed on the given circle as follows.
Steps of construction :
1. Join AE and bisect it. Let F be the mid-point of AE. 2. Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. Join AG.
AB and AG are the required tangents.
