Given, ABC is a right triangle in which `AB=6cm, BC=8cm,angleB=90^(@)and BD` is perpendicular to AC.
Steps of Construction :
1. Draw the line segments AB = 6 cm and BC= 8 cm perpendicular to each other. Join AC, thus `DeltaABC` is the right triangle.
2. Take mid-point F of BC as centre, draw a circle with radius 4 cm , passing through B,C and D.
3. Now, join AF and bisect it. Let mid-point of AF is O.
4. Take O as centre and OA as radius, fraw a circle which intersect the given circle (which passes through B,C and D) at B and M.
5. Now, join AB and AM, which are the required tangents.
Justification : Join FM and FB. Then, `angleAMF` is the angle lie in semicircle, so
`" "angleAMF=90^(@)`
`implies" "FMbotAM`
Since, FM is radius of the circle. So, AM has to be tangent of the circle with centre F. Similarly, AB is also a tangent to the circle with centre F.
