MCQ
In Fig, $\angle A C B=\angle C D A, A C=8 cm, A D=3 cm$, then $B D=$
- A$\frac{22}{3} cm$
- B$\frac{26}{3} cm$
- ✓$\frac{55}{3} cm$
- D$\frac{64}{3} cm$
✓
Answer
Correct option: C.
$\frac{55}{3} cm$
(C)$\frac{55}{3} cm$
In triangles $A C B$ and CDA, we find that
$
\angle A C B=\angle C D A\qquad[Given]
$
and
$
\angle C A B=\angle C A D
$
So, by using $A A$-criterion of similarity, we obtain
$\triangle A C B \sim \triangle A D C \Rightarrow \frac{A C}{A D}=\frac{A B}{A C} \Rightarrow \frac{8}{3}=\frac{A D+D B}{8} \Rightarrow \frac{64}{3}=3+D B \Rightarrow D B=\frac{55}{3} cm$
In triangles $A C B$ and CDA, we find that
$
\angle A C B=\angle C D A\qquad[Given]
$
and
$
\angle C A B=\angle C A D
$
So, by using $A A$-criterion of similarity, we obtain
$\triangle A C B \sim \triangle A D C \Rightarrow \frac{A C}{A D}=\frac{A B}{A C} \Rightarrow \frac{8}{3}=\frac{A D+D B}{8} \Rightarrow \frac{64}{3}=3+D B \Rightarrow D B=\frac{55}{3} cm$
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