Pg 222 and question no. 6 of ncert
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, and D is drawn. Construct the tangents from A to this circle. Give the justification of the construction.
while doing this question i facing problem in drawing the line BD
the ncert solutions on this site are also not helpfull(so please dont copy and paste the answer that is on this site)
D is the point where the line segment AC intersects the circle. Join BD.
As BD ⊥ AC, ∴∠BDC = 90°. Thus, if a circle is drawn through B, D, and C, BC will be its diameter. The centre E of this circle is the mid-point of BC.
The required tangents can be constructed on the given circle as follows.
1) Join AE and bisect it. Let F be the mid-point of AE.
2) Taking F as centre and FE as radius, draw a circle which will intersect the circle at point B and G. Join AG.
AB and AG are the required tangents.
Justification of construction: The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.
∠AGE is angle in the semi-circle.
∴ ∠AGE = 90° (angle in a semi-circle is a right angle)
⇒ EG ⊥ AG
Since EG is the radius of the circle, AG has to be a tangent of the circle.
But we have proved ∠B = 90°
⇒ AB ⊥ BE
Since BE is the radius of the circle, AB has to be a tangent of the circle.