In triangle ABC, ad bisectors BAC and CE are perpendicular to point O and intersect AB at point E. try to explain the relationship between de and DC and explain the reason

In triangle ABC, ad bisectors BAC and CE are perpendicular to point O and intersect AB at point E. try to explain the relationship between de and DC and explain the reason

Ad is the bisector of ∠ BAC, so ∠ Cao = ∠ EAO
∵CE⊥AD
∴∠AOC=∠AOE=90°
Ao is the common edge
∴△ACO≡△AEO
Then co = EO
∠COD=∠EOD=90°
Do is the common side
∴△CDO≡△EDO
So CD = De

In ABC, ad vertical BC, perpendicular foot D, CE vertical AB, perpendicular foot e, which means that the triangle BDE is similar to the triangle BAC

Ad vertical BC, CE vertical ab
Triangle abd is similar to triangle CBE,
AB：CB=BD：BE
AB：BD=CB：BE
Triangle BDE similar triangle bac

It is known that D. e is any two points in the triangle ABC. Try to explain the size relationship between ab + AC and BD + de + CE
connect BD.DE.EC To form an irregular quadrilateral,

As shown in the figure, the sum of the two sides of the triangle is greater than the third side
AB+AC＝AB+AF+FC＞BF+FB＞BG+GC＞BD+CE+EC

As shown in the figure, D and E are two internal points of △ ABC, which are ab + AC > BD + de + CE

It is proved that extending de and ed to f and g respectively, AF + Ag > FG (1) in △ AFG, FB + FD > BD (2) in △ BFD, eg + GC > EC (3) in △ EGC, ∵ FD + ed + eg = FG, ∵ ① + ② + ③, AF + FB + FD + FD + eg + GC + Ag > FG + BD + EC, namely AB + FD + eg + AC > FG + BD + EC, AB + AC > fg-fd-eg + BD + EC, ∵ AB + AC > BD + ed + EC