# in the given figure AB and AC are 2 tangents to a circle from an external point a DE is line segment touching the circle at F and intersecting AB at D and AC at E prove that AB =1/2 the perimeter of triangle ADE

Dear Student,

Tangents drawn from an external point to the circle are equal.
DB = DF ….(1)
EF = EC ….(2)
AB = AC ….(3)

Now,
AB = AC.
⇒ AD + DB = AE + EC
⇒ AD + DF = AE + EF  -------- ( Using (1) and (2) )

= AD + ( DF + FE ) + EA
= (AD + DF) + (FE + EA)
= 2 ( AD + DF ) ( AD + DF = AE + EF​ )
= 2 ( AD + DB ) --------  ( Using (1) )
= 2 AB

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• 6
Here is similar 1 :-

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) )
= 2 AQ
​​

Hope it helps ..!!!

• 3
thanksss
• 1
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