Geometry

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 $AZ=2ZC$ .

#Geometry

Easy Math Editor

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Comments

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.

Ayush G Rai - 4 years, 8 months ago

Why go for Menelaus when Thales' can do ?

Vishwash Kumar ΓΞΩ - 4 years, 7 months ago

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.