Construct a triangle, given the altitude, median, and angle bisector for a vertex. We are given that in a triangle, say $\triangle ABC$, the altitude is dropped from  A to the opposite side of the triangle. Also given is the median from A and it's the angle bisector.
With help of the above conditions, construct $\triangle ABC$.

I tried solving this problem. I know that if it is an equilateral or isosceles triangle the question can be easily done. But there should definitely be a general proof for any triangle.
Construct a triangle, given three distinct lengths of cevians from the
same vertex, say, $A$, the median $m_a$, the angle bisector $\beta_a$
and the altitude $h_a$, $h_a&lt;\beta_a&lt; m_a $.

Known relations between $m_a,\, \beta_a,\, h_a$
and the side lengths $a,b,c$ are:
\begin{align} 
2m_a^2&amp;=b^2+c^2-\tfrac12a^2
\tag{1}\label{1}
,\\
\beta_a^2&amp;=
bc\left(1-\frac{a^2}{(b+c)^2}\right)
\tag{2}\label{2}
,\\
h_a^2 &amp;= 
\frac{4b^2c^2-(b^2+c^2-a^2)^2}{4\,a^2}
\tag{3}\label{3}
.
\end{align}
Excluding $b,c$ from \eqref{1}-\eqref{3},
we get a quadratic expression in $a^2$
\begin{align} 
(a^2)^2-8\,(m_a^2-2h_a^2)\,a^2
&amp;+\frac{16(m_a^2-\beta_a^2)(\beta_a^2\,m_a^2-\beta_a^2\,h_a^2-m_a^2\,h_a^2)}{\beta_a^2-h_a^2}
=0
\tag{4}\label{4}
,
\end{align}
which gives the value of the side length $a$.
Equation \eqref{1} provides the value of $b^2+c^2$ in terms of $a$,
and \eqref{3} provides the value of $b^2c^2$ in terms of $a$,
which gives another quadratic equation with the roots $b^2,c^2$.
So the expressions for
the side length $a$ and
the other two side lengths
in terms of $a$ are
\begin{align} 
a_{1,2}&amp;=
2\sqrt{m_a^2-2h_a^2
\pm
(2h_a^2-\beta_a^2)\sqrt{\frac{m_a^2-h_a^2}{\beta_a^2-h_a^2}}}
\tag{5}\label{5}
,\\
b,c&amp;=
\tfrac12\sqrt{4m_a^2+a^2
\pm
4a\sqrt{m_a^2-h_a^2}}
\tag{6}\label{6}
.
\end{align}
Example
\begin{align} 
h_a&amp;=3
,\quad
\beta_a=4
,\quad
m_a=5
.
\end{align}
Eq. \eqref{4} gives two roots
\begin{align} 
a_1&amp;=
\tfrac27\,\sqrt{343+56\sqrt7} 
\approx 6.332
,\\
a_2&amp;=
\tfrac27\,\sqrt{343-56\sqrt7} 
\approx 3.988
,
\end{align}
corresponding
\begin{align}
b_1,c_1&amp;= 
\tfrac27\,\sqrt{392+14\sqrt7 
\pm 14\sqrt{343+56\sqrt7}}
\approx 
7.768649668. 3.113762020
,\\
b_2,c_2&amp;= 
\tfrac27\,\sqrt{392-14\sqrt7 
\pm 14\sqrt{343-56\sqrt7}}
\approx 
6.702893563, 3.608848334
.
\end{align}
Verification confirms
that triple $(a_1,b_1,c_1)$
indeed corresponds to
the triangle with given $h_a,\beta_a,m_a$:

Edit
As the
answer
to the follow-up
question
proves,
the only valid root is always $a_1$,
\begin{align} 
a_{1}&amp;=
2\sqrt{m_a^2-2h_a^2
+
(2h_a^2-\beta_a^2)\sqrt{\frac{m_a^2-h_a^2}{\beta_a^2-h_a^2}}}
.
\end{align}
In the indicated case it is given that the median $m_a$ from $A$ and the angle bisector $w_A$ at $A$ coincide. In this case the triangle is isosceles, and the altitude from $A$ coincides with the line $m_a=w_A$. It follows that the three occurring lengths coincide, and are equal to the height $h_a=|AM_a|$, where $M_a$ is the midpoint of $BC$ on the base of the triangle. When the given lengths $m_a$, $\beta_a$, and $h_a$ do not coincide there is no triangle having the desired properties.
On the other hand, when $m_a=\beta_a=h_a$, the given information does not determine the two points $B$, $C$ at equal distance from $M_a$ on the base.