Question

79. O paralelă la mediana AD a triunghiuluiABC întîlneşte respectiv laturile AB, AC, in E şi F. Să se demonstreze că: AE/AF=AB/AC ​

Answer 1

Explicație pas cu pas:

D ∈ BC, BD ≡ DC

M ∈ BC, ME || AD, E ∈ BA, ME ∩ AC = {F}

=> ΔACD ~ ΔFCM

$\frac{AC}{FC} = \frac{DC}{MC} \iff \frac{AC}{AC - FC} = \frac{DC}{DC - MC} \\ \implies \frac{AC}{AF} = \frac{DC}{DM}\iff \frac{AC}{AF} = \frac{BC}{2DM}$

=> ΔEBM ~ ΔABD

$\frac{EB}{AB} = \frac{BM}{BD} \iff \frac{EB - AB}{AB} = \frac{BM - BD}{BD} \\ \implies \frac{AE}{AB} = \frac{DM}{BD} \iff \frac{AE}{AB} = \frac{2DM}{BC}$

$\frac{AC}{AF} = \frac{BC}{2DM}\iff \frac{AF}{AC} = \frac{2DM}{BC}$

$\frac{AE}{AB} = \frac{AF}{AC} \iff \frac{AE}{AF} = \frac{AB}{AC}$

q.e.d.