1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 = ___________.

Open in App
Solution

In ∆ABC, sides AB and AC are extended to D and E, respectively such that AB = BD and AC = CE. Now, $\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{AB}+\mathrm{BD}}=\frac{\mathrm{AB}}{\mathrm{AB}+\mathrm{AB}}=\frac{\mathrm{AB}}{2\mathrm{AB}}=\frac{1}{2}.....\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AC}}{\mathrm{AE}}=\frac{\mathrm{AC}}{\mathrm{AC}+\mathrm{CE}}=\frac{\mathrm{AC}}{\mathrm{AC}+\mathrm{AC}}=\frac{\mathrm{AC}}{2\mathrm{AC}}=\frac{1}{2}.....\left(2\right)$ 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) $⇒\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{AE}}$ (If two triangles are similar, then their corresponding sides are proportional) $⇒\frac{1}{2}=\frac{6\mathrm{cm}}{\mathrm{DE}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{DE}=2×6=12\mathrm{cm}$ 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__.

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
Pythogoras Theorem
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program