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The lines $AB$, $AC$ and $DE$ are tangent to the circle with center $O$, the points $D$ and $E$ respectively belong to the segments $[AB]$ and $[AC]$.

$| AB | = l$ and $| OB | = | OC | = r$. enter image description here

what is the perimeter of the triangle $ADE$?

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up vote 4 down vote accepted

Do a drawing, and call $\,M\,$ to the tangency point of $\,DE\,$ with circle $\,O\,$: Since tangents from the same point exterior to the circle have equal lengths, we get:


Thus, the perimeter of $\,\Delta ADE\;$ is (spoiler!):

$$AB-DB +DE +AE=(l-DB)+(DM+ME) +(l-EC)=$$$$=l-\color{red}{DB}+\color{red}{DM}+\color{green}{ME}-\color{green}{EC}+l=2l$$

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Antonio , what is the value of DM and EM interms of l and r ? – Harish Kayarohanam Aug 20 '13 at 11:15
got it cleared , thanks – Harish Kayarohanam Aug 20 '13 at 11:16
But perimeter is AD + DE + AE . BUT AB-DB is AB , how can it account for permiter ? – Harish Kayarohanam Aug 20 '13 at 11:19
ya diagram makes this clear . I thought it was ADE circumscribing the circle with tangential points B, M ,C . Thanks . – Harish Kayarohanam Aug 20 '13 at 11:27
Well, the result is clear, and some geometry books even mention it as a theorem: with the given data, the area of triangle $\,\Delta ADE\;$ depends only on the length of (any of) the tangent from A to the circle, and equals twice this length. Of course, the length of cord $\,BC\,$ depends on where the vertex $\,A\,$ is wrt the circle and thus also on the length of the tangent lines, but as given you can see it all boils down to $\,l\,$ . – DonAntonio Aug 20 '13 at 11:38

lemma:Let c be a Excircle of $\Delta ADE$ and has tangency point M on $DE$,then $AD+DM$ is semiperimeter of $\Delta ADE$.

We have $DB=DM$ and $AB=l$.now with the above lemma,the perimeter is $2l$.

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Edited, thanks. – R Salimi Aug 20 '13 at 13:47

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