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 $| OB | = | OC | = r$.
what is the perimeter of the triangle $ADE$?
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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:
$$DB=DM\;,\;\;EM=EC$$
Thus, the perimeter of $\,\Delta ADE\;$ is (spoiler!):
$$AB-DB +DE +AE=(l-DB)+(DM+ME) +(l-EC)=$$$$=l-\color{red}{DB}+\color{red}{DM}+\color{green}{ME}-\color{green}{EC}+l=2l$$
DonAntonio
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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
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got it cleared , thanks – Harish Kayarohanam Aug 20 '13 at 11:16
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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
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D is point on extended AB and E is a point on extended AE ? – Harish Kayarohanam Aug 20 '13 at 11:21
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i will add a draw – user52413 Aug 20 '13 at 11:21
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yes without diagram it confuses. – Harish Kayarohanam Aug 20 '13 at 11:22
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@HarishKayarohanam: $$DB=DM\implies -DB+DM=0;.;;ME=EC\implies ME-EC=0\ldots$$ so we do not care what these quantities are: they cancel! – DonAntonio Aug 20 '13 at 11:22
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But for triangle perimeter it is AD + DE + AE = (AB+BD) + (DM + ME) + (EC+ AC) = 2AB + 2DM + 2EC .where comes cancelling – Harish Kayarohanam Aug 20 '13 at 11:26
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@HarishKayarohanam, did you read my whole answer? For it you have to put your mouse point over the gray area to see what's written there (spoiler)... – DonAntonio Aug 20 '13 at 11:26
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1ya 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
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Ok. Anytime @HarishKayarohanam – DonAntonio Aug 20 '13 at 11:28
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Thanks for handling me with patience . I have upvoted your answer . +1 – Harish Kayarohanam Aug 20 '13 at 11:29
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i 'd like to understand why the area is the same if $R$ and $BC$ don't change. – user52413 Aug 20 '13 at 11:30
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1Well, 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
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thx but it missing p=f(L/r) – user52413 Aug 20 '13 at 11:46
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@user52413, I've no idea what you wrote above: what is p, what is f...? – DonAntonio Aug 20 '13 at 11:55
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@DonAntonio p for perimeter et f for function. – user52413 Aug 20 '13 at 12:06
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Well, then $P_{\Delta ADE}=2l=f(l,r);$ . This function is constant in $,r,$ , though I'm not sure what that $,L/r;$ there means... – DonAntonio Aug 20 '13 at 12:07
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i was looking for something better than trigonometry $l=rtan(\widehat{AOC}) , and P=2r*tan(\widehat{AOC})$ – user52413 Aug 20 '13 at 12:23
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My demonstration only uses basic euclidean geometry, though I wouldn't say that is "better" than trigonometry...and I can't understand the point of your formulas: do you want to add a new unknown $,\angle AOC;$ ?! – DonAntonio Aug 20 '13 at 12:28
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i don't know $AD/AB=DE/BC$ so $ADBC/AB=DE$ and if $BC/DE=1$ then $AD =DE$ and $P =3AD$ – user52413 Aug 20 '13 at 13:14
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@user52413, I'm trying to understand what you mean but I can't: isn't my answer satisfactory? If it is not, why? Now you bring me some geometry stuff which I'm not sure how it is related with the answer I wrote to you...Is there anything you don't understand in my answer? – DonAntonio Aug 20 '13 at 13:17
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nevermind, i'll ask an another one. – user52413 Aug 20 '13 at 14:02
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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 $2l$.
R Salimi
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