Giải thích các bước giải:
Bài 1:
Ta có :
$\dfrac{x}{2}=\dfrac{y}{3}$
$\to \dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2+y^2}{4+9}=\dfrac{208}{13}=16$
$\to\dfrac{x}{2}=\dfrac{y}{3}=4\to x=8,y=12\to z=15$
Hoặc $\dfrac{x}{2}=\dfrac{y}{3}=-4\to x=-8,y=-12\to z=-15$
Bài 2:
a.Ta có $AN=NC, MN=ND, \widehat{ANM}=\widehat{CND}\to\Delta AMN=\Delta CDN(c.g.c)$
$\to MA=CD$ mà $MA=MB\to MB=CD$
b.Gọi $AE\cap BC=F$
$\to\widehat{BEC}=\widehat{BEF}+\widehat{FEC}=\widehat{BAE}+\widehat{ABE}+\widehat{EAC}+\widehat{ECA}=\widehat{BAC}+\widehat{ABE}+\widehat{ACE}>\widehat{BAC}$