In the triangle ABC, BD and CE divide ⊥ ABC, ⊥ ACB, AF ⊥ BD in F and Ag ⊥ CE in G equally. If AB = 6, BC = 12 and AC = 10, find the length of FG Such as the title It's urgent

In the triangle ABC, BD and CE divide ⊥ ABC, ⊥ ACB, AF ⊥ BD in F and Ag ⊥ CE in G equally. If AB = 6, BC = 12 and AC = 10, find the length of FG Such as the title It's urgent

It is proved that extending AG, crossing BC at point m, extending AF, crossing BC at point n
∵ BD bisection ∠ ABC, an ⊥ BD
Easy syndrome △ Abf ≌ △ NBF
∴AF=FN,BA=BN
Similarly, Ag = GM, CA = cm
Ψ GF is the median of △ anm
∴FG=1/2MN=1/2(BN+CM-BC)=1/2(6+10-12)=2

As shown in the figure, it is known that BD and CE are the heights of AC and BC sides of △ ABC respectively, and G and F are the midpoint of BC and de respectively

It is proved that: as shown in the figure, connecting eg, DG, ∵ BD and CE are the heights of AC and BC sides of △ ABC respectively, point G is the midpoint of BC, ∵ DG = eg = 12bc, ∵ point F is the midpoint of De, ∵ GF ⊥ De