What's the measure of the side BC of the triangle below?

Question

For reference:

In the triangle $ABC$, the heights $BE$ and $CD$ are drawn. If $AC \cdot CE= 88$ and $AB \cdot BD= 108$. Calculate $BC$ (Answer:$14$)

My progress:

$AC\cdot CE = 88 \rightarrow b \cdot n =88 \\ AB \cdot BD = 108 \rightarrow c \cdot BD = 108\\ \triangle ABE\sim \triangle ACD \implies\\ \frac{AB}{AC}=\frac{BE}{CD}=\frac{AE}{AD}\rightarrow \frac{c}{b}=\frac{h}{p}=\frac{m}{AD}\\ \triangle CDB:: p^2+BD^2 = a^2\\ \triangle CDA: DA^2+p^2=b^2\\ a^2-BD^2 =b^2-AD^2 \\ \text{T.Poncelet} \triangle ACD-ED\\ m\cdot CH \cdot BD = n \cdot HD \cdot c\implies\\ \text{T.Poncelet} \triangle ABE-DC\\ AD \cdot BH \cdot n =BD \cdot EH \cdot b$

but I can't equate...

Answer 1

Notice that $ACFD$ and $ABFE$ are concyclic. So by the power of the points $C$ and $B$ we have $$CF\cdot CB= AC\cdot CE = 108$$ and $$BF\cdot CB= BD\cdot BA = 88$$ so $$BC ^2 = BF\cdot CB + CF\cdot CB = 88+108 = 196$$ and thus $BC = 14$.

Answer 2

As $ \small BDEC$ is cyclic, $ \small HE \cdot BH = CH \cdot HD$ and as $ \small ADHE$ is cyclic, $ \small BH \cdot BE = BD \cdot AB, CH \cdot CD = CE \cdot AC$. We will use them in our work below.

$$ \small \begin {aligned} BC^2 & = BE^2 + CE^2 \\ & = (BH+HE) \cdot BE + CE^2 \\ & = BH \cdot BE + HE \cdot (BH+HE) + CE^2 \\ & = BH \cdot BE + HE \cdot BH + HE^2 + CE^2 \\ & = BH \cdot BE + HE \cdot BH + CH^2 \\ & = BD \cdot AB + CH \cdot (HD + CH) \\ & = BD \cdot AB + CH \cdot CD \\ & = BD \cdot AB + CE \cdot AC \\ \end {aligned} $$