Geometry

Let XX be the midpoint of the side AB AB of triangle ABCABC. Let YY be the midpoint of CXCX. Let BYBY cut ACAC at ZZ. Prove that AZ=2ZCAZ=2ZC.

Note by Subham Subian
3 years ago

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We don't have to use cyclic properties.We extend the line AYAY to meet BCBC at K.K.We apply Menelaus theorem for BCX\triangle BCX where AKAK is the transverse.So we get BAAX×XYYC×KCKB=1.\dfrac{BA}{AX}\times\dfrac{XY}{YC}\times\dfrac{KC}{KB}=-1.Since BABA is twice of AXAX and XY=YC.XY=YC.We get KCKB=12.\dfrac{KC}{KB}=\dfrac{1}{2}.We don't have care about the minus sign as it only to show AA does not lie between BX.BX.Next we apply Ceva's theorem for ABC.\triangle ABC.We get AXBX×BKKC×ZCAZ=1.\dfrac{AX}{BX}\times\dfrac{BK}{KC}\times\dfrac{ZC}{AZ}=1.We know that AX=BXAX=BX and BK=2KC.BK=2KC.So we get ZCAZ=12.\dfrac{ZC}{AZ}=\dfrac{1}{2}.Therefore,AZ=2ZC.\boxed{AZ=2ZC}.Hence proved.

Ayush G Rai - 3 years ago

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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)=ar(ABC)4ar(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=xx and ZC=yy
Now,
\frac {x}{x+y} * ar (ABC) = ar(ABZ) \(\implies \frac {x}{x+y}* ar (ABC) = \frac{ar (ABC)}{2} + ar (CYA)
    xx+yar(ABC)=ar(ABC2+xx+yar(ABC)4\implies \frac {x}{x+y} * ar (ABC) = \frac {ar (ABC}{2} + \frac{x}{x+y} * \frac {ar (ABC)}{4}
    xx+y[ar(ABC)ar(CYA)]=ar(ABC)12\implies \frac {x}{x+y}[ ar(ABC) - ar (CYA)] = ar(ABC)* \frac {1}{2}
    xx+y[34ar(CYAB)=ar(ABC)2\implies \frac {x}{x+y}[\frac {3}{4} * ar (CYAB) = \frac{ar (ABC)}{2}
    xx+y[34ar(ABC)\implies \frac {x}{x+y} * [\frac {3}{4}* ar (ABC)
    xx+y=23\implies \frac {x}{x+y} = \frac {2}{3}
    x=2y\implies x= 2y
Hence, proved.

Yatin Khanna - 3 years ago

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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.

Yatin Khanna - 3 years ago

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

Yatin Khanna - 3 years ago

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

Yatin Khanna - 3 years ago

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

vishwash kumar - 3 years ago

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

Yatin Khanna - 3 years ago

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

Subham Subian - 3 years ago

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

abhishek alva - 3 years ago

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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

Subham Subian - 3 years ago

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