Question

In $\Delta ABC,AB=BC=6cm$. If a circle is drawn with centre at B and radius 2 cm which intersects AB and BC at E and D respectively, then:

• A. $\frac{AB}{EB}=\frac{BC}{ED}$
• B. $\frac{AB}{EB}=\frac{AC}{ED}$
• C. $\Delta ABC\sim \Delta EBD$
• D. $\Delta BAC\sim \Delta EBD$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{AB}{EB}=\frac{AC}{ED}$
C $\Delta ABC\sim \Delta EBD$

Given: AB = BC = 6 cm
BE = BD = 2 cm (radii of same circle)

In $\Delta ABC$ and $\Delta EBD$,

$\frac{BE}{BA}=\frac{BD}{BC}=\frac{2}{6}=\frac{1}{3}$

$\angle DBE=\angle ABC$ (common in both triangles)

$\therefore \Delta ABC\sim \Delta EBD$ (by SAS similarity criteria)

$\Rightarrow \frac{AB}{EB}=\frac{BC}{BD}=\frac{AC}{ED}$
(corresponding sides of similar triangles are in same ratio)