Question

D and E are the point on side AB and AC respectively of a triangle ABC . Such that DE parallel to BC and divides Triangle ABC into two parts , equal in area , find BD/AB

Answer 1

Answer:

$\frac{BD}{AB}=\frac{2-\sqrt{2}}{2}$

Step-by-step explanation:

It is given that,

D and E are the point on side AB and AC respectively of a triangle ABC . Such that DE parallel to BC and divides Triangle ABC into two parts , equal in area

Therefore we can write,

$\frac{ar(ADE)}{ar(ABC)}=\frac{1}{2}$

(since DE divides Triangle ABC into two parts)

To find BD/AB

$\frac{ar(ADE)}{ar(ABC)}=\frac{AD}{AB}^{2}$

$\frac{ar(ADE)}{ar(ABC)}=\frac{1}{2}$  

$\frac{AD}{AB}^{2}=1/2$

Therefore,

$\frac{AD}{AB}=\frac{1}{\sqrt{2} }$

$1-\frac{AD}{AB}=1-\frac{1}{\sqrt{2} }$

$\frac{AB}{AB}-\frac{AD}{AB}=\frac{\sqrt{2} }{\sqrt{2}}-\frac{1}{\sqrt{2} }$

$\frac{BD}{AB}=\frac{\sqrt{2}-1}{\sqrt{2} }$

$\frac{BD}{AB}=\frac{2-\sqrt{2}}{2}$

Answer 2

As triangle ADE  ~ ABC

AD/AB = DE/BC

8/8+12 = DE/BC

8/20 = DE/BC

2/5 = DE/BC

BC = 5/2 DE

AD = 6 cm; BD = 9 cm

AE = 8 cm  and CE = 12 cm

Now we have:

AD/BD = 6/9 = 2/3

AE/CE = 8/12 = 2/3

So AD/BD = AE/CE

We know that if a line intersects any two sides of a triangle in equal ratio then the line is parallel to the third side.

DE = BC