Question

De is drawn parallel to the base bc of a triangle abc, meeting ab at d and ac at e. Ab/bd=4 and ce=2cm, find ae

Answer 1

Answer:  6

Step-by-step explanation:

Here ABC is a triangle,

In which $D\inAB$ and $E\inAc$

Also, $DE\parallel BC$

Thus, $\angle ADE\cong \angle ABC$  ( by The alternative interior angle theorem)

Similarly,  $\angle AED\cong \angle AEB$

Thus, By AA similarity postulate,

$\triangle ADE\sim \triangle ABC$

By the property of similar triangles,

$\frac{AD}{AB} = \frac{AE}{AC}$

⇒ $-\frac{AD}{AB} = -\frac{AE}{AC}$ ( By multiplying -1 on both sides )

⇒ $1-\frac{AD}{AB} = 1-\frac{AE}{AC}$  ( By adding 1 on both sides)

⇒ $\frac{AB-AD}{AB} = \frac{AC-AE}{AC}$

⇒ $\frac{BD}{AB} = \frac{CE}{AC}$

⇒  $\frac{1}{4} = \frac{2}{AC}$ ( because $\frac{AB}{BD} = 4$ and CE=2 )

⇒  $AC = 8$

But,  AE = AC - CE = 8 - 2 = 6

Thus, AE = 6 cm

Answer 2

Answer:

The answer is (C) AB/BD=AC/CE

Step-by-step explanation: