Question

in figure,BC is a diameter of circle with centre M .PA si tangent at A from P which is a point on line BC.AD perpendicular to BC .Prove that DP2=BP*CP-BD*CD

Answer 1

Answer:

Step-by-step explanation:

By tangent sec theorem.

1. Ap^2 = Bp*Cp....... 1

2. in tri ADP by pytho th

AP^2 = AD^2+DP^2

DP^2=AP^-AD^2.....2

3. From 1 and 2

DP^2=BP*CP-AD^2.....3

TRIANGLE BAD AND TRIANGLE ACD are similar right angle test (both are right angled triangke and are inside the right angle triangle BAC which is opposite to diameter hence BAC is 90°

Therefor,

4. AD/CD = BD/AD.... Sides of similar traingle are in proportion

AD^2=BD*CD....4

FROM 3 AND 4

DP^2=BP*CP-BD*CD

Answer 2

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