# In a right-angled triangle ABC which has right angle at B. A circle is drawn with AB as diameter intersecting the hypotenuse at P. Prove that the tangent to the circle at P bisects BC.

Answer

Verified

335.1k+ views

**Hint**: We first draw the information given for us. We can see that side BC is tangent to the circle from a point where tangent P is also passing. We use the fact that angles opposite to equal sides are also equal, sum of angles in a straight line is $ {{180}^{o}} $ to prove the required result.

**:**

__Complete step-by-step answer__Given that we have a right-angled triangle ABC whose right angle is at vertex B and we have a circle drawn with AB as diameter intersecting the hypotenuse at P and a tangent is drawn to circle at point P. We need to prove that this tangent bisects the side BC.

Let us draw a figure so that all the information can be viewed better. We assume the tangent of circle at point P intersects side BC at point ‘F’. According to the problem we need to prove BF=FC.

We join the points B and P. We know that the normal of any circle passes through the center of the circle and lies perpendicular to the tangents.

So, we can see that the side BC is perpendicular to the diameter AB. So, side BC satisfies the condition of becoming tangent.

We can see that the tangents BC and PF passes through the point F. This makes lines FB and FP are tangents drawn externally from the point F.

We know that if the lengths of the tangents drawn from the external point to the point of contact on the circle are equal. By using this point we get BF = PF ---(1).

We know that sides opposite to the equal sides has equal angles.

So, we get $ \angle PBC=\angle BPF $ ---(2).

From triangle APB, we can see that $ \angle APB={{90}^{o}} $ ---(3).

We know that the sum of angles lie on a straight line constitutes $ {{180}^{o}} $ . Using this fact on side AC, we have

$ \angle APB+\angle CPB={{180}^{o}} $ .

From equation (3), we have $ \angle APB={{90}^{o}} $ .

$ {{90}^{o}}+\angle CPB={{180}^{o}} $ .

$ \angle CPB={{90}^{o}} $ ---(4).

From triangle BPC we know that sum of angles in a triangle is $ {{180}^{o}} $ ,

$ \angle PBC+\angle CPB+\angle BCP={{180}^{o}} $ .

From equation (4),

$ \angle PBC+{{90}^{0}}+\angle BCP={{180}^{o}} $ .

$ \angle PBC+\angle BCP={{90}^{o}} $ .

From (2), we got

$ \angle BPF+\angle BCP={{90}^{o}} $ ---(5).

We know that $ \angle CPB={{90}^{o}} $ ,

$ \angle CPF+\angle BPF={{90}^{o}} $ .

$ \angle BPF={{90}^{o}}-\angle CPF $ ---(6).

Substituting equation (6) in equation (5) we get,

$ {{90}^{o}}-\angle CPF+\angle BCP={{90}^{0}} $ .

$ \angle BCP={{90}^{0}}-{{90}^{o}}+\angle CPF $ .

$ \angle BCP=\angle CPF $ ---(7).

The sides opposite to angles $ \angle BCP,\angle CPF $ are PF, CF which are also equal.

So, we got PF=CF ---(8)

From equations (1) and (8), we got BF=CP.

∴ We proved that the tangent PF bisects the side BC.

**Note**: We should not directly take the value of the angle $ \angle PFB $ as $ {{90}^{o}} $ , as the value of the angle will not be the same. We should know that the normal of the circle will always be perpendicular to the tangent and passes through the center of the circle. Here the angle $ \angle APB $ is $ {{90}^{o}} $ , as the angle between the lines connecting both ends of a semicircle is $ {{90}^{o}} $ .

Last updated date: 26th Sep 2023

•

Total views: 335.1k

•

Views today: 6.35k

Recently Updated Pages

What do you mean by public facilities

Please Write an Essay on Disaster Management

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is the IUPAC name of CH3CH CH COOH A 2Butenoic class 11 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

The dimensions of potential gradient are A MLT 3A 1 class 11 physics CBSE

Define electric potential and write down its dimen class 9 physics CBSE

Why is the electric field perpendicular to the equipotential class 12 physics CBSE