Question

Answer 1

AE/AB=CD/CB

AF/AC=BD/CB

Do đó: AE/AB+AF/AC=1

Answer 2

Xét ΔBAC có DE//AC

nên \(\frac{BE}{BA}=\frac{BD}{BC}\)

=>\(1-\frac{BE}{BA}=1-\frac{BD}{BC}\)

=>\(\frac{AE}{AB}=\frac{CD}{CB}\)

Xét ΔCAB có DF//AB

nên \(\frac{CD}{CB}=\frac{CF}{CA}\)

=>\(1-\frac{CD}{CB}=1-\frac{CF}{CA}\)

=>\(\frac{AF}{AC}=\frac{BD}{BC}\)

\(\frac{AE}{AB}+\frac{AF}{AC}\)

\(=\frac{BD}{BC}+\frac{CD}{BC}=\frac{BC}{BC}=1\)