Question

In $△ABC$, AD is the median. Which of these conditions should be satisfied to make $\Delta ADB$ and $\Delta ADC$ similar triangles?

• A. $\angle A={90}^{\circ }$
• B. $AB=AC$
• C. $\angle B=\angle C$
• D. (b) or (c) above.

Answer 1

The correct option is D (b) or (c) above.

In the triangles $△ADB$ and $\Delta ADC$,
AD is the common side
BD = DC (Since D is the midpoint of BC)

Hence ratio of corresponding sides is:
$\frac{AD}{AD}=1$,
$\frac{BD}{DC}=1$,
$\frac{AB}{AC}$

For the triangles to be similar, all three ratios should be equal.
$⟹\frac{AB}{AC}=1$
$AB=AC$
Since opposite angles to equal sides in isoceles triangle are equal.
$⟹\angle B=\angle C$

Hence one of the conditions is enough as the other can be inferred from either Option B or Option C.