Let \(X\) be the midpoint of the side \( AB\) of triangle \(ABC\). Let \(Y\) be the midpoint of \(CX\). Let \(BY\) cut \(AC\) at \(Z\). Prove that \(AZ=2ZC\).

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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## Comments

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

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

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

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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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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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