A circle touches the side Bc of a triangle ABC at a point P and touches AB and AC when produced at Q and R respectively. Show that

  1. AQ = 1/2 ( perimeter of triangle ABC)

Hi!
Here is the answer to your query.
 
 
Tangents drawn from an external point to the circle are equal.
BP = BQ  ….(1)
CP = CR  ….(2)
AQ = AR  ….(3)
Now, AQ = AR
⇒ AB + BQ = AC + CR
⇒ AB + BP = AC + CP  (Using (1) and (2))
 
Perimeter of ∆ABC = AB + BC + CA
  = AB + (BP + PC) + AC
  = (AB + BP) + (PC + AC)
  = 2(AB + BP)  ( AB + BP = AC + CP)
  = 2(AB + BQ)  (Using (1))
  = 2AQ

Cheers!

  • 474

 we know that tangents drawn from an exterior poin 2 a circle are equal

AQ = AR     1

BP = BQ     2

CP = CR     3

perimeter of ABC = AB+BC+AC

= AB+BP+CP+AC

=AB+BQ+CR+AC   [Using 2 & 3]

= AQ+AR

=2AQ  [using 1]

hence, AQ = 1/2(perimeter of triangle ABC)

thumbs up plz... cheers!!!

  • 111

Did nt understand after this step

AB + BQ + CR + AC

after this

how it this comes : AQ + AR

  • -30
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