Question

A triangle ABC is right angled at B..D and E are distinct points on
AC such that BD and BE trisect AC. If BD = a, BE = b and AC = 3
a2+2
c then the value of
is equal to:
22​

Answer 1

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

Hi, , hope it is helpful ⬆️