Question

va rog imi trebuie repede dau coronita la toti care imi raspund

Answer 1

[AB[=[AC]
[AB]/2=[AC]/2⇒[EB]=[DC]

ΔBEC ≡ ΔCDB caz LUL
[EB]=[DC]
BC latura comuna
m(∡EBC)=m(∡DCB)
⇒[BD]=[CE]

Answer 2

[tex]\it [AB]\equiv [AC] \Longrightarrow \Delta ABC -isoscel \Longrightarrow \widehat{ABC} \equiv \widehat{ACB}\ \ \ \ (1) [/tex]

[tex]\it [AB]\equiv [AC] \Longrightarrow AB=AC \Longrightarrow \dfrac{AB}{2} =\dfrac{AC}{2} \Longrightarrow EB = DC \Longrightarrow \\\;\\ \\\;\\ \Longrightarrow [EB] \equiv [DC] \ \ \ \ (2)[/tex]

[tex]Vom \ compara\ \Delta BCE\ cu\ \Delta CBD: [/tex]

[tex]\it [BC] \equiv [CB] \ (latura\ comun\breve{a})\ \ \ (*) \\\;\\ (1) \Rightarrow \widehat{EBC} \equiv \widehat{DCB}\ \ \ \ (**) [/tex]

[tex]\it [EB] \equiv [DC] \ \ \ \ \ (***) \\\;\\ Din\ (*),\ (**),\ (***) \stackrel{LUL}{\Longrightarrow} \Delta BCE \equiv \Delta CBD \Longrightarrow [CE]\equiv [BD] \Longrightarrow \\\;\\ \Longrightarrow [BD]\equiv [CE] \ \ \ \ \ \ \ [q.\ e.\ d.][/tex]