Question

how we can find segmant of triangle

Answer 1

ΔACE
because ΔACE
and
ΔBDE
ΔBDE
are similar, we have:
|BE|+|CE|
|AC|
=
|BD|
|DE|
|BE|+|CE||AC|=|BD||DE|
We also have:
|AD|+|BD|
|AC|
=
|BE|
|DE|
|AD|+|BD||AC|=|BE||DE|
and so:
|BE|+|CE|
|BD|
=
|AD|+|BD|
|BE|
|BE|+|CE||BD|=|AD|+|BD||BE|
Using
|AC|=|AD|=8
|AC|=|AD|=8
and
|CE|=|DE|
|CE|=|DE|
this becomes:
|BE|(|BE|+|CE|)=|BD|(8+|BD|)
|BE|(|BE|+|CE|)=|BD|(8+|BD|)
We also have that
|BD
|
2
+|CE
|
2
=|BE
|
2
|BD|2+|CE|2=|BE|2
, so we arrive at a final equation in
|BE|,|CE|
|BE|,|CE|
. But
|CE|
|CE|
is (reasonably) any value, so there is no fixed value to
|BE|
|BE|
.