Q8 of 24 Page 296

AD, BE and CF are the medians of ΔABC. If AD = BC, then let us prove that the angle between two medians is 90°.


Given: AD = BC


As we know, if G is centroid,


Then AD = AG


⇒ AG = BC


⇒ if AG = 2 then GD = 1 ….(1)


Also, BC = 2 (BG = AG)


BD = DC = 1 (median) ….(2)


From (1) and (2)


BD = DC = GD = 1


⇒ ∠GDB = 90° and ∠GDC = 90°


Because, BD = GD


⇒ ∠GBD = ∠DGB = x (let) (isosceles triangle)


Hence,


x + x + 90° = 180°


2x = 90°


⇒ x = 45°


Similarly,


Because, DC = DG


⇒ ∠GCD = ∠DGC = y (let) (isosceles triangle)


Hence,


y + y + 90° = 180°


2y = 90°


⇒ y = 45°


And x + y = 45° + 45°


⇒ ∠BGC = 90°


i.e. angle between median is 90° .


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