Question

In the following figure, AB be the chord of a circle with centre' P.
Tangents at points A and B intersect at point C. Prove that:
AM^2 = PM × CM.​

Answer 1

Answer:

Step-by-step explanation:

proof:

let∠APM=∅

so∠PAM=90-∅

now,

       ∠PAM+∠MAC=90°

      90-∅+∠MAC=90°

      ∅=∠MAC=∠APM

in ΔAMP and ΔAMC

∠AMC=∠AMP             [90°]

∠MAC=APM               {∠MAC=∅}

ΔMAC≈ΔAPM             [AA similarity]

AM/PM=CM/AM          [BPT]

AM²=CM×PM

HENCE PROVED