In which of the following cases, DE ǁ AB where D and E lie on the sides AC and BC of ∆ABC respectively?
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If a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same ratio.
As DE∥BC
∠ADE=∠ABC
∠AED=∠ACB
So by AAA △AED∼△ACB
Hence
AC
AE
=
CB
ED
∴
AC
3.2
=
5
2
⇒AC=8cm
AC=8cm=AE+EC=3.2+EC
EC=8−3.2=4.8cm
Also
AB
AD
=
CB
ED
∴
AB
2.4
=
5
2
AB=6cm=AD+DB=2.4+DB
⇒DB=3.6cm
So BD=3.6cm and CE=4.8cm
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