Question

In the figure, AD = AE and $A{D}^2=BD\times EC$. Then triangles ABD and CAE are not similar triangles.

• A. True
• B. False

Answer 1

The correct option is B

False

In the given figure,

AD = AE

$A{D}^2=BD\times EC$

In $\Delta ADC,AD=AE$

$\therefore \angle ADE=\angle AED$ (Angles opposite to equal sides)

But $\angle ADE+\angle ADB=\angle AED+\angle AEC={180}^{\circ }$

$\therefore \angle ADB=\angle AEC$

$A{D}^2=BD\times EC$

$\frac{AD}{BD}=\frac{EC}{AD}$

Therefore $\Delta ABD\sim \Delta CAE$

Hence the given statement is false.