Question

Următoarea problemă, vă rog !
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$\it \hat In\ \Delta ABC\ avem:\\ \\ \hat A=\hat B=72^o,\ AD-\ bisectoare\ pentru\ \hat A,\ D\in AC,\ AD=\sqrt5\ cm.\\ \\ Determina\c{\it t}i\ perimetrul \ triunghiului.$

Answer 1

ΔABC - isoscel, AC = BC

ΔABD -isoscel, AB = AD = √5 cm

ΔADC -isoscel, AD= CD = √5 cm

$\Delta ABC\sim\ \Delta BDA \Rightarrow\dfrac{AB}{BD}=\dfrac{BC}{AD}\\ \\ \dfrac{\sqrt5}{BD}=\dfrac{BC}{\sqrt5} \Rightarrow BD\cdot BC=\sqrt5\cdot\sqrt5\\ \\ (BC-\sqrt5)\cdot BC=5 \Rightarrow BC^2-\sqrt5BC-6=0$

$\it \Rightarrow BC=\dfrac{\sqrt5+5}{2}\ cm \Rightarrow AC=\dfrac{\sqrt5+5}{2}\ cm\\ \\ \\ \mathcal{P} =AB+BC+CA = \sqrt5+\dfrac{\sqrt5+5}{2}+\dfrac{\sqrt5+5}{2}=2\sqrt5+5\ cm$