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
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$\begingroup$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.
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