≦=≤≥lt； & gt+-×÷/∫∮∝∞

∑∪∩∈∵∴⊥∠⌒⊙≌∽√πΩ

^△α

Analytic geometry can be used

Take point B as the coordinate origin, BC as the X axis, B as the Y axis perpendicular to BC, and let the side length of a regular triangle be

a

therefore

B(0，0)

，A(a/2

,√3/2

a

)，D(2/3

a

,0),

C(a，0)

AD:y=-3√3

X+2√3

a

tan∠ECB=tan(∠ADB-60 )=(

tan∠ADB-tan 60)/( 1+tan∠ADB

*tan60)

=(3√3-√3)/( 1+9)=√3/5

that is

K of EC =-√ 3/5

, so

EEC:

y=-√3/5(X-a)

A.D., European Community

Simultaneous intersection of two linear equations

E

(9/ 14

a

,

√3/ 14

answer

So it is

K=√3/9

And k of AD =-3 √ 3.

Mutually negative reciprocal

, so AD⊥BE,

∠AEB=90