Question

In the given figure, the line segment XY is Parallel to AC of ABC and it

divides the triangle into two parts of equal areas. Prove that

AX/AB=√2-1/√2

Answer 1

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Answer 2

Answer:

Step-by-step explanation:Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices

•AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.

SOLUTION:

GIVEN:

XY || AC

ar(∆BXY) = ar(quad.XYCA)

ar(∆ABC) = 2ar(∆BXY).................(1)

Now, XY || AC & BA is a transversal .

∠BXY = ∠BAC…………………(2)

[corresponding angles]

In ∆ BAC & ∆BXY,

∠BXY = ∠BAC     (Proved above)

∠XBY = ∠ABC      (Common)

∆ BAC ~ ∆BXY    

[By AA similarity criterion]

ar(∆BAC) / ar(∆BXY) = BA² /BX²

[The ratio of the area of two similar triangles is equal to the ratio of the square of any two corresponding sides]

2ar(∆BXY) / ar(∆BXY) = BA² /BX²

2 = BA² /BX²

BA² = 2 × BX²

BA =√2 × BX²

BA = √2 BX

BA = √2 (BA - AX)      [ BX = BA - AX ]

BA = √2BA - √2AX

√2AX = √2BA - BA

√2AX = BA (√2- 1)

AX / AB = (√2- 1) / √2

Hence, proved.

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