Question

$△$ ABC is an acute angled triangle. DE is drawn parallel to BC as shown. Which of the following are always true?

i) $△ABC\sim △$ ADE

ii) $\frac{AD}{BD}=\frac{AE}{EC}$

iii) $DE=\frac{BC}{2}$

• A. Only (i)
• B. (i) and (ii) only
• C. (i), (ii) and (iii)
• D. (ii) and (iii) only

Answer 1

The correct option is B

(i) and (ii) only

Consider $△ADE$ and $△ABC$

Since DE || BC,
$\frac{AD}{BD}=\frac{AE}{EC}$
(by basic proportionality theorem)

$\angle BAC=\angle DAE$ (common in both the triangles)

$\therefore △ADE\sim △ABC$ (by SAS similarity criterion)

$\Rightarrow \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$
(corresponding sides of similar triangles are in same ratio)

Now, $DE=\frac{BC}{2}$ only if

$\Rightarrow \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}=\frac{1}{2}$

i.e. D and E should be the midpoints of AB and AC respectively.
So this may not be true always.