Question

$AB$ is the diameter of a circle of a circle, $CD$ is a chord parallel to $AB$ and $2CD=AB$. The tangent at $B$ meets the line $AC$ produced at $E$. Prove that $AE=2AB$.

Answer 1

Given ,

$2CD=AB$ & $AB||CD$

$AD=DB=OC=OD=CD=radius$

$AB=2R.$

As $,AB||CD\Rightarrow AC=OD=r$

Triangle ACD will be equilateral Triangle & $\angle CAD={60}^0$

$\tan A=\frac{BE}{BA}=\sqrt{3}\Rightarrow BE=\sqrt{3}\left(2r\right).$

$A{E}^2=B{A}^2+B{E}^2$

$=(2r{)}^2+(\sqrt{3}\times 2r{)}^2$

$=4r^2+3\times 4r^2$

$=16r^2$

$AE=4r=2\left(2r\right)=2\left(AB\right).$

$AE=2AB.$ Hence Proved.