Question

$ABC$ is an isosceles triangle such that $AB=AC$ .$D$ is the mid point of $AC$ . A circle is drawn taking $BD$ as diameter which intersects $AB$ at point $E$ .then $AC=3AE$ .

• A. True
• B. False

Answer 1

The correct option is B False

$ABC$ is an isosceles triangle in which $AB=AC,$

$D$ is mid-point of $AC.$ A circle with diameter $BD$ intersect $AB$ in $E.$

$AD$ is a tangent and $AEB$ is a secant

$\therefore$ $A{D}^2=AE\times AB$

$\Rightarrow$ ${\left(\frac{1}{2}AC\right)}^2=AE\times AC$ [ As $AB=AC$ and $AD=\frac{1}{2}AC$ ]

$\Rightarrow$ $\frac{1}{4}AC=AE$

$\Rightarrow$ $AC=4AE$