If AB and CD are the two tangents of a circle intersect each other at P, then prove that AP × BP = CP × DP.
Answers
Answer:
Mark me as brainliest
Step-by-step explanation:
If two chords of a circle intersect each other internally or externally, then area of rectangle contained by the segment of one chord is equal to the area of rectangle contained by the segment of other chord.
Using above theorem,
PA × PB = PC × PD …(1)
(i) Given: AB = 4 cm
BP = 5 cm
PD = 3 cm
∵ AP = AB + BP
⇒ AP = 4 cm + 5 cm
⇒ AP = 9 cm
Putting the values in (1),
9 cm × 5 cm = PC × 3 cm
⇒ PC = 15 cm
∵ PC = CD + DP
⇒ 15 cm = CD + 3 cm
⇒ CD = 15 cm – 3 cm
⇒ CD = 12 cm
Hence, CD = 12 cm
(ii) Given: BP = 3 cm
CP = 6 cm
CD = 2 cm
∵ CP = CD + DP
⇒ DP = CP – CD
⇒ DP = 6 cm – 2 cm
⇒ DP = 4 cm
Putting the values in (1),
PA × 3 cm = PC × PD
⇒ PA × 3 cm = 6 cm × 4 cm
⇒ PA = 8 cm
∵ AP = AB + BP
⇒ 8 cm = AB + 3 cm
⇒ AB = 8 cm – 3 cm
⇒ AB = 5 cm
Hence, AB = 5 cm