a line DE is drawn parallel to base BC,meeting AB in D and AC at C.If AB/BD=4 and CE=2cm.Find length of AE
Answers
Answer:
Here ABC is a triangle,
In which D\inABD\inAB and E\inAcE\inAc
Also, DE\parallel BCDE∥BC
Thus, \angle ADE\cong \angle ABC∠ADE≅∠ABC ( by The alternative interior angle theorem)
Similarly, \angle AED\cong \angle AEB∠AED≅∠AEB
Thus, By AA similarity postulate,
\triangle ADE\sim \triangle ABC△ADE∼△ABC
By the property of similar triangles,
\frac{AD}{AB} = \frac{AE}{AC}
AB
AD
=
AC
AE
⇒ -\frac{AD}{AB} = -\frac{AE}{AC}−
AB
AD
=−
AC
AE
( By multiplying -1 on both sides )
⇒ 1-\frac{AD}{AB} = 1-\frac{AE}{AC}1−
AB
AD
=1−
AC
AE
( By adding 1 on both sides)
⇒ \frac{AB-AD}{AB} = \frac{AC-AE}{AC}
AB
AB−AD
=
AC
AC−AE
⇒ \frac{BD}{AB} = \frac{CE}{AC}
AB
BD
=
AC
CE
⇒ \frac{1}{4} = \frac{2}{AC}
4
1
=
AC
2
( because \frac{AB}{BD} = 4
BD
AB
=4 and CE=2 )
⇒ AC = 8AC=8
But, AE = AC - CE = 8 - 2 = 6
Thus, AE = 6 cm