Question

Let ABC be the right triangle in which AB = 6cm, BC = 8cm, $\angle B = 90^\circ ,$ BD is perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangent from A to this circle.

Answer 1

Hint: To solve above question first we will construct right triangle $\Delta ABC$ as D is perpendicular on an AC and touches the circle. So, BC must be the diameter of the circle and from the midpoint of BC we can get the centre of the circle whose radius will be $\dfrac{1}{2}BC = 4cm$. Then we will construct longest to it

Complete step by step solution:
Rough sketch

Now first we will construct $\Delta ABC$ it is given that $BC = 8cm, AB = 6cm,\angle B = 90^\circ $
i. Construct triangle, $\Delta ABC$ with given measurements
ii. Draw perpendicular bisector of line BC

iii. Let the line intersect BC at point E. Now, E is the midpoint of BC.
iv. Taking E as a centre and BE as radius, draw a circle.

v. Join AE and bisect it, let it intersect AE at point M.
vi. Now taking M as centre and AM as radius draw a circle.

vii. Points where this second circle touches the first circle are points of contacts.
viii. These points of contact are B and Q.
ix. A and B are already joined, join A and Q.
x. AB and AQ are required tangents of the circle.

So, AB and AQ are the required tangents.

Note: 1. While solving any problem on construction we should always make a rough figure to get an idea.
2. Every measurement should be precise so that we can avoid error in final construction.