≦=≤≥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