Question

Let abcd be a rectangle. the perpendicular bisector of line segment bd intersect ab and bc in point e and f respectively. let m and n are midpoints of cd and ad respectively. prove that line fm and en are perpendicular.

Answer 1

Given :- Let ABCD be a rectangle. The perpendicular bisector of line segment BD intersects lines AB and BC in points E and F, respectively. Let M and N be midpoints of line segments CD and AD, respectively.

To Prove :- Lines FM and EN are perpendicular.

Concept used :-

• Coordinate geometry .
• Slope of line A(x1,y1) and B(y2,y2) = (y2 - y1)/(x2 - x1) .
• Mid points of A(x1,y1) and B(y2,y2) = (x1 + x2)/2 , (y1 + y2)/2 .
• Equation of a line :- y - y1 = m(x - x1) .
• when m1 * m2 = (-1) , lines are ⟂ .

Solution :- { Refer to image. }

Let ABCD is a rectangle, where

• A coordinates as (0, 2b).
• B coordinates as (2a, 2b).
• C coordinates as (2a, 0).
• D A coordinates at origin (0, 0).
• AB = DC , AD = BC .

since , M and N are mid - points of CD and AD

→ M = (2a + 0)/2 , (0 + 0)/2 = (a , 0)

→ N = (0 + 0)/2 , ), (2b + 0)/2 = (0, b)

now, perpendicular bisector of line segment BD intersects lines AB and BC in points E and F, respectively.

so,

→ coordinates of E = (a² - b²/a , 2b)

→ coordinates of F = {2a , - (a² - b²)/b}

using, mid - point and slope formula we get, { Refer to image. }

→ Equation of line FM (in form of y = mx + c) :- y = -(a² - b²/ab)x + (a² - b²/b)

→ Equation of line EN :- y = (ab/a² - b²)x + b .

then,

→ Slope of line FM = -(a² - b²/ab) = m1

→ Slope of line EN = (ab/a² - b²) = m2

checking,

→ m1 * m2

→ -(a² - b²/ab) * (ab/a² - b²)

→ (-1) .

since product of slope is equal to (-1), therefore, we can conclude that, FM ⟂ EN .

{ Excellent Question .}

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