Question

Point B ∈ |AC| so that AB:BC=2:1. Point D ∈ |AB| so that AD:DB=3:2. Find AD:DC

Answer 1

Answer:

5:4

Step-by-step explanation:

If point B divides the segment AC in the ratio 2:1, then

AB=2x units and BC=x units.

If point D divides the segment AB in the ratio 3:2, then

AD=3y units and DB=2y units.

Since AD+DB=AB, then

$3y+2y=2x\\ \\5y=2x\\ \\y=\dfrac{2}{5}x$

Now,

$AD=3y\\ \\DC=DB+BC=2y+x=2y+\dfrac{2}{5}y=\dfrac{12}{5}y$

So,

$AD:DC=3y:\dfrac{12}{5}y=15:12=5:4$

Answer 2

Answer:

AD:DC=6:9

Step-by-step explanation:

We know that:

AB:BC=2:1

AD:DB=3:2

We can conclude that:

AB+BC=AC

Then:

AB=2/3AC

BC=1/3AC

AD+DB=AB

Then

AD=3/5AB

DB=2/5AB

From the above we can replace:

AD=(3/5)(2/3AC)=6/15AC

On the other hand:

DC= DB+BC

DC=2/5AB+1/3AC

In terms of AC

DC=((2/5)(2/3AC))+1/3AC=4/15AC+1/3AC

DC=27/45AC=9/15AC

From:

AD=6/15AC

DC=9/15AC

we can say that:

AD:DC=6:9