in triangle abc D and e are points on AB and ac such that De parallel to BC ad equal to 2 cm, ab is equal to 6 CM AC equal to 9 cm find ae
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f 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\parallel BCDE∥BC
\angle ADE=\angle ABC∠ADE=∠ABC
\angle AED=\angle ACB∠AED=∠ACB
So by AAAAAA \triangle AED\sim \triangle ACB△AED∼△ACB
Hence \dfrac{AE}{AC}=\dfrac{ED}{CB}ACAE=CBED
\therefore\dfrac{3.2}{AC}=\dfrac{2}{5}∴AC3.2=52
\Rightarrow AC=8cm⇒AC=8cm
AC=8cm=AE+EC=3.2+ECAC=8cm=AE+EC=3.2+EC
EC=8-3.2=4.8cmEC=8−3.2=4.8cm
Also \dfrac{AD}{AB}=\dfrac{ED}{CB}ABAD=CBED
\therefore \dfrac{2.4}{AB}=\dfrac{2}{5}∴AB2.4=52
AB=6cm=AD+DB=2.4+DBAB=6cm=AD+DB=2.4+DB
\Rightarrow DB=3.6cm⇒DB=3.6cm
So BD=3.6cmBD=3.6cm and CE=4.8cmCE=4.8cm

Step-by-step explanation:
hope u understand
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