Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Construction Procedure:
The given circle’s tangents can be constructed as follows
1. Draw the line segment BC of measure 8cm.
2. At the point B measure the angle 90°, such that ∠ B = 90°.
3. Draw an arc with a measure of 6cm taking B as centre.
4. The point where the arc intersects the ray be A.
5. Now join the line AC.
6. Therefore, the required triangle is ABC.
7. Now, to the line BC draw a perpendicular bisector and mark the midpoint as E.
8. Draw a circle taking E as centre and BE or EC as radius.
9. Join A to the circle’s midpoint E.
10. Now, to the line AE draw the perpendicular bisector taking the midpoint as M.
11. Draw a circle taking M as Centre and ME or AM as radius.
12. At the points B and Q, this circle intersects the previous circle.
13. Now join the points Q and A.
14. Therefore, the required tangents are AB and AQ.
Justification:
By proving that AG and AB are the tangents to the circle the construction can be justified.
From the construction above, join EQ.
As we know that an angle in a semi-circle is a right angle and ∠AQE is an angle in the semi-circle, therefore
∠AQE = 90°
⇒ EQ⊥ AQ
AQ has to be a tangent of the circle, since EQ is the radius of the circle. In the similar way, ∠B = 90°
⇒ AB ⊥ BE
AB has to be a tangent of the circle, since BE is the radius of the circle.
As a result, the above construction is justified.