Mathematics
In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.
Prove that : AC = 3AF.
Answer
By mid-point theorem,
The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.
By converse of mid-point theorem,
The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.
Draw DG || BF
In △ ADG,
⇒ DG || EF (Since, DG || BF)
Since, E is the mid-point of AD,
∴ F is the mid-point of AG (By converse of mid-point theorem)
∴ AF = FG ………(1)
In △ BCF,
⇒ DG || BF
As, AD is the median, D is the mid-point of BC.
∴ G is the mid-point of FC (By converse of mid-point theorem)
∴ FG = GC ………(2)
From equations (1) and (2), we get :
⇒ AF = FG = GC.
From figure,
⇒ AC = AF + FG + GC
⇒ AC = AF + AF + AF
⇒ AC = 3AF.
Hence, proved that AC = 3AF.
Related Questions
In a trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that :
AB + DC = 2EF.
In △ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
A school designing a triangular garden △ABC. To construct a walking path inside the garden, the gardener marks the mid-points of two sides: point D is the mid-point of side AB and point E is the midpoint of side AC. The path DE is drawn to connect these mid-points. The length of side BC of the triangular garden is 12 m, AB = 10 m and AC = 10 m.
Based on the above information answer the following:
(i) What is the length of path DE?
(ii) Assign a special name to Quadrilateral BCED and find its perimeter.