Question

ABC is an isosceles triangle in which AB = AC A circle through B touches side AC at its mid point D and intersects AB at P, then AP will be-

• A. 2AD
• B. $\frac{1}{2}AB$
• C. $\frac{1}{4}AB$
• D. $\frac{1}{2}AC$

Answer 1

The correct option is C $\frac{1}{4}AB$

Now D is a midpoint so,
$AD=\frac{AC}{2}=\frac{AB}{2}$
According to theorem-
$AP\times AB=A{D}^2$
$AP\times AB=\frac{A{B}^2}{4}$
$AP=\frac{1}{4}AB$