Question

In ∆ABC, sides AB and AC are extended to D and E respectively, such that AB = BD and AC = CE. If BC = 6 cm, then DE = ___________.

Answer 1

In ∆ABC, sides AB and AC are extended to D and E, respectively such that AB = BD and AC = CE.



Now,

$$\begin{gathered} \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{AB}+\mathrm{BD}}=\frac{\mathrm{AB}}{\mathrm{AB}+\mathrm{AB}}=\frac{\mathrm{AB}}{2\mathrm{AB}}=\frac{1}{2}.....\left(1\right) \\ \frac{\mathrm{AC}}{\mathrm{AE}}=\frac{\mathrm{AC}}{\mathrm{AC}+\mathrm{CE}}=\frac{\mathrm{AC}}{\mathrm{AC}+\mathrm{AC}}=\frac{\mathrm{AC}}{2\mathrm{AC}}=\frac{1}{2}.....\left(2\right) \end{gathered}$$

From (1) and (2)

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

In ∆ABC and ∆ADE,

$\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}$ (Proved)

∠A = ∠A (Common)

∴ ∆ABC $~$ ∆ADE (SAS Similarity)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{AE}}$ (If two triangles are similar, then their corresponding sides are proportional)

$$\begin{gathered} \Rightarrow \frac{1}{2}=\frac{6\mathrm{cm}}{\mathrm{DE}} \\ \Rightarrow \mathrm{DE}=2\times 6=12\mathrm{cm} \end{gathered}$$

In ∆ABC, sides AB and AC are extended to D and E respectively, such that AB = BD and AC = CE. If BC = 6 cm, then DE = __12 cm__.