$H$ is the orthocentre of an acuteangled triangle $ABC$. A point $E$ is taken on the line segment $CH$ such that $ABE$ is a rightangled triangle. Prove that the area of the triangle $ABE$ is the geometric mean of the areas of triangles $ABC$ and $ABH$.
Let $CH$ meet $AB$ at $G$ .Now the given question is equivalent to proving $GE^2$ =$GH$×$GC$
Notice that $GE$ is also the geometric mean of $AG$ and $GB$.( because it is right angled and $GE$ is altitude).So, $GE^2$=$AG$×$GB$
So we need to prove $AG$×$GB$=$GH$×$GC$, which is true because triangle $AGH$ is similar to triangle $CGB$.
Here is what I did: Extend CH to meet AB at D, By $Trignometry$ we $AH = 2R cos A$ similarly $BH = 2R cos B$ , WE know $[ABC]= 1/2 ab sin C$ and $[ABH]=2R^2cos A cos B sin C $,
So $\sqrt([ABH]*[ABC])= R sin C \sqrt (ab cos A cos B)$ = c /2* \sqrt(AD*DB)= c/2* DE = [ABE]
Hence proved
