Question

In figure the line segment XY is parallel to side AC of Triangle ABC and it divides the triangle in to two parts of equal areas. Find the ratio AX/AB​

Answer 1

Given:

• a triangle ABC

• line segment XY is parallel to side AC

To Find:

ratio AX : AB

Solution:

XY||AC (given)

so,

BXY = A (corresponding angles)

BYX = C (corresponding angles)

$∴$ ∆ABC ≈∆XBY (AAA similarly criteria)

So,

$⟹\sf{\frac{ar(ABC )}{ar(XBY)} = (\frac{AB}{XB} )^{2} .......(1)}$

$⟹\sf{ar(ABC) = 2ar(XBY)}$

$⟹\tt{\frac{ar(ABC)}{ar(XBY)} = \frac{2}{1} .....(2)}$

From ( 1 ) and ( 2 ),

$⟹\large\tt{( \frac{AB}{XB} )^{2} = \frac{2}{1} \: \: i.e. \: \frac{AB}{XB} = \frac{ \sqrt{2} }{1}}$

$⟹\large\tt{\frac{XB}{AB} = \frac{1}{ \sqrt{2} }}$

$⟹\large\sf{1 - \frac{XB}{AB} = 1 - \frac{1}{\sqrt{2} } }$

$⟹\large\sf{ \frac{AB - XB}{AB} = \frac{ \sqrt{2} - 1}{ \sqrt{2} }}$

$⟹\large\sf{ \frac{AX}{AB} = \frac{ \sqrt{2} - 1 }{ \sqrt{2} } = \frac{2 - \sqrt{2} }{2}}$

Hence,

$\Large\tt\red{ \frac{AX}{AB} = \frac{2 - \sqrt{2} }{2}}$

Answer 2

Answer:

Given:

• a triangle ABC

• line segment XY is parallel to side AC

________________________

To Find:

ratio AX : AB

________________________

Solution:

XY||AC (given)

so,

BXY = A (corresponding angles)

BYX = C (corresponding angles)

∴ ∆ABC ≈∆XBY (AAA similarly criteria)

So,

⟹ ar(ABC ) / ar(XBY) = (AB / XB)^2 .......(1)

⟹ ar(ABC) = 2ar(XBY)

⟹ ar(ABC) / ar(XBY) = 2 / 1 ........(2)

From ( 1 ) and ( 2 ) ,

⟹ (AB / XB)^(2) = 2 / 1

⟹ AB / XB = √2 / 1

⟹ XB / AB = 1 / √2

⟹ 1 - XB / AB = 1 - 1 / √2

⟹ AB / AB - XB / AB

⟹ √2 / √2 - 1 / √2

⟹ AX / AB = ( 2 - √2 ) / 2