Question: Prove that $DO$, $DB$ and $DE$ are AM, GM and HM of $a$ and $b$ in the given figure.
It is given that $OA$ us the radius of this semi-circle.
I have proved $DO$ as AM of $a$ and $b$ simply by using the fact $DO$ is the radius of this semi-circle and will be equal to the half of the diameter $(a + b)$
And $DB$ can be proved as GM by first finding out $BO$ and the applying Pythagoras theorem in the triangle $DBO$.
But I'm stuck at proving $DE$ as HM of $a$ and $b$. I have tried it by first proving the triangles $DBO$ and $DEB$ similar and then using corresponding sides and area theorem but unable to conclude HM.
Kindly help.
$$R=AO=BO=OD=\frac{a+b}{2}$$ $$\angle ADC=90^{\circ} \Rightarrow DB^2=AB \cdot BC \Rightarrow DB=\sqrt{ab}$$
$\triangle ODB \sim \triangle BDE \Rightarrow \frac{DB}{DO}=\frac{DE}{DB}$ $$DB^2=DE \cdot DO \Rightarrow DE=\frac{DB^2}{DO}=\frac {ab}{\frac{a+b}{2}}=\frac{2ab}{a+b}$$