Question

in triangle abc and triangle xyz,if ab/xz=bc/xy=ac/yz then triangle abc and triangle xyz​

Answer 1

Answer:

If ABC and XYZ are two triangles such that AB : BC = XY: YZ and the angles B and Y are supplementary angles, how do I prove that {Area ABC/Area XYZ} = (AB/XY) ^2?

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If △ABC and △XYZ are two triangles such that AB:BC=XY:YZ and the angles B and Y are supplementary, how do I prove that Area△ABCArea△XYZ=(ABXY)2?

Construction: Draw a line l perpendicular to BC and passing through B as well a line m perpendicular to YZ and passing through Y as shown in the figure below.

AB:BC=XY:YZ⇒ABXY=BCYZ.

Let DB and WY be the projections of AB and XY on lines l and m respectively.

⇒DB and WY are the heights of △ABC and △XYZ respectively.

It is given that ∠ABC and ∠XYZ are supplementary.

Let ∠ABC=θ⇒∠XYZ=180o−θ.

⇒∠DBA=90o−θ, and,

∠XYW=180o−θ−90o=90o−θ.

⇒∠DBA=∠XYW.

⇒cos∠DBA=cos∠XYW.

⇒DBAB=WYXY⇒DBWY=ABXY=BCYZ.

⇒Area△ABCArea△XYZ=BC×DBYZ×WY=(BCYZ)(DBWY)

=(ABXY)(ABXY)=(ABXY)2.