Given
In isosceles ΔABC the segment BD (with D∈ AC ) is the median to the base AC.
perimeter of ΔABC is 50 meters, and the perimeter of ΔABD is 40 meters
Find out the value of the BD.
To proof
In the isosceles triangle the median is also become the altitude.
First we prove that
ΔABD is congurent to Δ CBD
In ΔABD and Δ CBD
(1) AB = BC
( As given in the question ΔABC isaisosceles triangle thus two sides of this triangle are equal i.e AB = BC .)
(2) BD = BD
( common side of the triangle)
(3) AD = DC
( BD is a median i.e it divided AC in two equal parts i.e AD = DC )
Thus by using the SSS congurence property.
[tex]ABD\cong CBD[/tex]
Thus
perimeter of ΔABD = perimeter of Δ CBD = 40 meter
( by corresponding part of the congurent triangle)
let the length of the median (BD) = x
Thus the equation becomes
perimeter of ΔABD + perimeter of ΔCBD - 2 ( length of the median BD ) = perimeter of Δ ABC
putting the value in the above equation
we get
40 + 40 - 2x = 50
80 - 2x = 50
30 = 2x
x = 15 meters
thus the length of the median BD is 15 meters.
Hence proved
