Question:
In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and ∠AOC = 45°. Then, ∆OAC and ∆ODB are
(a) equilateral and similar
(b) equilateral but not similar
(c) isosceles and similar
(d) isosceles but not similar
Solution:
(c) isosceles and similar
In ∆AOC and ∆ODB, we have:
$\angle A O C=\angle D O B \quad$ (Vertically opposite angles)
and $\angle O A C=\angle O D B \quad$ (Angles in the same segment)
Therefore, by AA similarity theorem, we conclude that $\triangle A O C \sim \triangle D O B$.
$\Rightarrow \frac{O C}{O B}=\frac{O A}{O D}=\frac{A C}{B D}$
Now, $O B=O D$
$\Rightarrow \frac{O C}{O A}=\frac{O B}{O D}=1$
$\Rightarrow O C=O A$
Hence, $\triangle \mathrm{OAC}$ and $\triangle \mathrm{ODB}$ are isosceles and similar.