Q:

How to prove that the altitudes of the triangle are concurrent

Given a triangle of sides $A$ $B$ $C$ .how to prove that altitudes from sides $A$ $B$ $C$ are concurrent . There is a theorem called ceva's theorem.but i dont know how to use that theorem in this problem

A:

The question is slightly unclear. My way to interprete the request is: "Deliver a proof for the concurrence of the heights in a triangle by using the Theorem of Ceva". Here it is.
Let $\Delta ABC$ be a triangle, and let $D$, $E$, $F$ be on the sides $BC$, $CA$, $AB$, so that $AD$, $BE$, $CF$ are respectively perpendicular on these sides. In a picture:

The we have (without considering signs)
$$
\frac{DB}{DC}
=
\frac{DB}{DA}
\cdot
\frac{DA}{DC}
=
\frac{\cot B}{\cot C}\ .
$$
Now we build the unsigned product
$$
\frac{DB}{DC}\cdot
\frac{EC}{EA}\cdot
\frac{FA}{FB}
=
\frac{\cot B}{\cot C}\cdot
\frac{\cot C}{\cot A}\cdot
\frac{\cot A}{\cot B}\cdot
=
1\ .
$$
Let us now consider the signs. If $\Delta ABC$....

has all angles $<90^\circ$, then each fraction above has negative sign, so the signed product is $-1$.  
has an angle $=90^\circ$, say the angle in $A$, then the heights are concurrent in $A$. This case is clear. (And the above computation does not really make sense.)
has an angle $>90^\circ$, say the angle in $A$, then the heights from $B,C$ have the feet $E,F$ outside the side segments $CA$, $AB$, so the corresponding proportions have positive sign, the third proportion is negative.

We obtain (in the two unclear cases) the signed product
$$
\frac{DB}{DC}\cdot
\frac{EC}{EA}\cdot
\frac{FA}{FB}
=-1\ .
$$
The (reciprocal of the) Theorem of Ceva insures now $AD$, $BE$, $CF$ concurrent.

