# 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

Please follow the solution to the above given query:

**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) )

Perimeter of ∆ ADE = AD + DE + EA

= AD + ( DF + FE ) + EA

= (AD + DF) + (FE + EA)

= 2 ( AD + DF ) ( AD + DF = AE + EF )

= 2 ( AD + DB ) -------- ( Using (1) )

= 2 AB

DB = DF ….(1)

EF = EC ….(2)

AB = AC ….(3)

Now,

AB = AC.

⇒ AD + DB = AE + EC

⇒ AD + DF = AE + EF -------- ( Using (1) and (2) )

Perimeter of ∆ ADE = AD + DE + EA

= AD + ( DF + FE ) + EA

= (AD + DF) + (FE + EA)

= 2 ( AD + DF ) ( AD + DF = AE + EF )

= 2 ( AD + DB ) -------- ( Using (1) )

= 2 AB

$\Rightarrow \mathrm{Perimeter}\mathrm{of}\u25b3\mathrm{ADE}=2\mathrm{AB}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=\frac{\mathrm{Perimeter}\mathrm{of}\u25b3\mathrm{ADE}}{2}$

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