Răspuns:
[tex]\boxed{\boxed{\bf \angle AFB = 130^{\circ}}}[/tex]
Explicație pas cu pas:
[tex]\bf ABC-isoscel \implies \angle B=\angle C \; si \; AB \equiv AC\;(1)[/tex]
[tex]\bf \angle A + \angle B + \angle C =180^{\circ}\; (2)[/tex]
[tex]\bf Fie \; AB = AC =x\; (3)[/tex]
[tex]\bf din \: (1),(2),(3) \implies x+x+20^{\circ}=180^{\circ}[/tex]
[tex]\bf \implies 2x=180^{\circ}-20^{\circ}=160^{\circ}|:2[/tex]
[tex]\bf \implies AB=AC=80^{\circ}[/tex]
[tex]\bf BE-bisectoarea\; \angle B \implies \angle ABE=\angle CBE=\dfrac{\angle B}{2}[/tex]
[tex]\bf \implies \angle ABE=\angle CBE =\dfrac{80^{\circ}}{2} =40^{\circ}[/tex]
[tex]\bf AD-inaltime \;dar \; \triangle ABC -isoscel[/tex]
[tex]\bf \implies AD-bisectoare,mediana,inaltime,mediatoare[/tex]
[tex]\bf AD-bisectoare \implies \angle BAD=\angle CAD=\dfrac{\angle A}{2}[/tex]
[tex]\bf \implies \angle BAD=\angle CAD=\dfrac{20^{\circ}}{2} =10^{\circ}[/tex]
[tex]\bf In \; \triangle ABF \; avem:[/tex]
[tex]\bf \angle BAD + \angle ABE + \angle AFB=180^{\circ}[/tex]
[tex]\bf \implies \angle AFB=180^{\circ}-\angle BAD - \angle ABE[/tex]
[tex]\bf \implies \angle AFB = 180^{\circ}-40^{\circ}-10^{\circ}=130^{\circ}[/tex]
Bafta! :)
#copaceibrainly
