In RT ABC, C 90 points E on AC BE bisected ABC ED vertically bisected AB at D if AC 9 finds the valu

Because ed bisects ab perpendicularly, ABE is an isosceles triangle, angle a=angle abe, and because be bisects angle abc, so angle abe = angle cbe, so angle abc = 2 * angle a, ae = be

Also considering the right triangle abc, the angle A + the angle abc = 90, so the angle cbe = angle A = 30, so in the cbe, be=2*ec, in summary, ae = be=2 * ec, at the same time ae + ec = ac = 9, so ae = 6

Be bisected abc, cbe= abe, ed bisected vertically ab to d, ea=eb, a= abe, cbe=30°, be=2ec, i.e., ae=2ec, while ae+ec=ac=9, ae=6

So the answer is: 6

If BAC is an acute angle, as shown in Figure 1:

The perpendicular bisector of ab is de, ae=be, ed ab, ad=1ab, ae=5, tan aed=3

4，∴sin∠aed=3

5，∴ad=ae?sin∠aed=3，∴ab=6，∴be+ce=ae+ce=ac=ab=6；

If the bac is obtuse, as shown in Figure 2:

The same can be obtained: be+ce=16

So the answer is: 6 or 16

Proof that in abc, acb=90°, ac=bc abc is an isosceles right triangle, abc=45° the bisector of abc intersects ac at the point d

abe=45°/2

In a right-angled triangle AEB.

tan∠abe=ae/eb

ae=eb*tan∠abe

tan∠abe=tan45°/2

[1-cos45°) 1+cos45°)] denotes square root) ae=eb*tan abe

eb*(√2-1)

√2-1)be

Select a ab=bf

Proof: BAC = 90°, AD BC

bad+∠abc=∠c+∠abc=90°∴∠bad=∠c

ef‖ac∠c=∠efb

efb=∠eab

abe=∠fbe，be=be

abe≌△bfe

ab=bfd is not possible unless c=30 degrees.

No, bfe= c= bae, but can't be sure that it is equal to abe!!

The extension cord of the be is crossed over the extension cord of the cd to f

CE is the bisector of BCD.

bce=∠fce

ab∥cd∠f=∠fba

Be is the bisector of ABC.

abf=∠fbc

fbc=∠f

and ce=ce

fce≌△bce

ef=be，bc=fc

def= aeb, ef=be, f= fba aeb def

ae=ed