Question

in the adjoing figure m<ABC= 57°M is the centre of the circle and line BC is a tangent seg BP is the diameter find​

Answer 1

Answer:

Step-by-step explanation:

Given: In the given figure BC is the diameter of the circle with centre M,PA is the tangent at A from P which is a point on line BC and AD⊥BC

To Prove:

DP  

2

=BP×CP−BD×CD

Construction: Join seg AB and seg AC

Proof: In △ABC,∠BAC=90  

∘

 (angle subtended by diameter in semi circle)

seg AD⊥ side BC ....(Given)

∴ By property of geometric mean,

AD  

2

=BD×CD ... (i)

Ray PA is a tangent and PB is a secant

∴ By tangent secant theorem,

PA  

2

=BP×CP ... (ii)

In right angled △PAD, the Pythagoras theorem,

PA  

2

=DP  

2

+AD  

2

 

∴DP  

2

=PA  

2

−AD  

2

 .... (iii)

From (i), (ii) and (iii), we get

DP  

2

=BP×CP−BD×CD

please mark as a brainliest