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Prove that if median drawn from $A$ and prepedicular drawn from $B$ and bisector drawn from $C$ intersect in the same point in the triangle $ABC$ then the triangle has three equal sides.

My attempt:I just tried to use Ceva's theorem but it failed.Also the angles don't give sth special.

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Let $BH$ be the altitude, $CL$ the angle bisector and $AM$ be the median. Ceva says that $$\dfrac{CH}{HA}\cdot\dfrac{AL}{LB}=1.$$ But $\dfrac{AL}{LB} = \dfrac{b}{a}$ and $\dfrac{CH}{HA} = \dfrac{a\cos\gamma}{c\cos\alpha}$, so we have that $$b\cos\gamma = c\cos\alpha.$$ This is equivalent to $\dfrac{a^2+b^2-c^2}{2a} = \dfrac{b^2+c^2-a^2}{2b}$ or $ba^2+b^3-bc^2-ab^2-ac^2+a^3 = 0.$ Ultimately, this is equivalent to: $$c^2 = \dfrac{a^2b-ab^2+a^3+b^3}{a+b}.$$

This means that the original assertion is false: Any triangle satisfying the above property will also satisfy the give requirement.

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