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# 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 $$AZ=2ZC$$.

Note by Subham Subian
1 year, 3 months ago

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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.

- 1 year, 3 months ago

Why go for Menelaus when Thales' can do ?

- 1 year, 2 months ago

- 1 year, 2 months ago

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

- 1 year, 2 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.

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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

- 1 year, 3 months ago

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

- 1 year, 3 months ago