We don't have to use cyclic properties.We extend the line $AY$ to meet $BC$ at $K.$We apply Menelaus theorem for $\triangle BCX$ where $AK$ is the transverse.So we get $\dfrac{BA}{AX}\times\dfrac{XY}{YC}\times\dfrac{KC}{KB}=-1.$Since $BA$ is twice of $AX$ and $XY=YC.$We get $\dfrac{KC}{KB}=\dfrac{1}{2}.$We don't have care about the minus sign as it only to show $A$ does not lie between $BX.$Next we apply Ceva's theorem for $\triangle ABC.$We get $\dfrac{AX}{BX}\times\dfrac{BK}{KC}\times\dfrac{ZC}{AZ}=1.$We know that $AX=BX$ and $BK=2KC.$So we get $\dfrac{ZC}{AZ}=\dfrac{1}{2}.$Therefore,$\boxed{AZ=2ZC}.$Hence proved.

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TopNewestWe don't have to use cyclic properties.We extend the line $AY$ to meet $BC$ at $K.$We apply Menelaus theorem for $\triangle BCX$ where $AK$ is the transverse.So we get $\dfrac{BA}{AX}\times\dfrac{XY}{YC}\times\dfrac{KC}{KB}=-1.$Since $BA$ is twice of $AX$ and $XY=YC.$We get $\dfrac{KC}{KB}=\dfrac{1}{2}.$We don't have care about the minus sign as it only to show $A$ does not lie between $BX.$Next we apply Ceva's theorem for $\triangle ABC.$We get $\dfrac{AX}{BX}\times\dfrac{BK}{KC}\times\dfrac{ZC}{AZ}=1.$We know that $AX=BX$ and $BK=2KC.$So we get $\dfrac{ZC}{AZ}=\dfrac{1}{2}.$Therefore,$\boxed{AZ=2ZC}.$Hence proved.

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Why go for Menelaus when Thales' can do ?

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Proof:

Note that $ar(AXY)=ar (BXY)=ar(BYC)=ar(AYC)=\frac{ar(ABC)}{4}$

Since, median divides a triangle into 2 parts of equal areas.

Next, put AZ=$x$ and ZC=$y$

Now,

\frac {x}{x+y} * ar (ABC) = ar(ABZ) \(\implies \frac {x}{x+y}* ar (ABC) = \frac{ar (ABC)}{2} + ar (CYA)

$\implies \frac {x}{x+y} * ar (ABC) = \frac {ar (ABC}{2} + \frac{x}{x+y} * \frac {ar (ABC)}{4}$

$\implies \frac {x}{x+y}[ ar(ABC) - ar (CYA)] = ar(ABC)* \frac {1}{2}$

$\implies \frac {x}{x+y}[\frac {3}{4} * ar (CYAB) = \frac{ar (ABC)}{2}$

$\implies \frac {x}{x+y} * [\frac {3}{4}* ar (ABC)$

$\implies \frac {x}{x+y} = \frac {2}{3}$

$\implies x= 2y$

Hence, proved.

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Its not z, its x, unable to edit.

And forgive me for the brackets not being closed and other mistakes that I am unable to edit.

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I was unable to solve it using my limited knowledge of cyclic quads.

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There is no 3/4 in 4th last line.

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This question can easily be solved using midpoint theorem and similarity rules as above.

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Seriously, BC cut AC at Z; wont that make Z coincide with C and consequently CZ will be 0.

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sorry my bad .i corrected it plss try to solve it and give mme the solution i will be gratefull

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you can use the cevas theorem and the menuluas theorem to prove this

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and can use properties of cyclic quadrilaterals beacause my teacher said that it can be solve by using the properties of cyclic quadrilaterals

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