D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm, Prove that BC = 5/2 DE.
Answers
Answered by
47
SOLUTION :
Given : AD = 8 cm,DB= 12 cm, AE = 6 cm, and CE = 9cm
AB = AD + DB = 8 + 12 = 20 cm
AB = 20 cm
AC = AE + EC
AC = 6 + 9 = 15 cm
AC = 15 cm
In ΔADE & ΔABC
AD/AB = 8/20 = 2/5
AE/AC = 6/15 = 2/5
AD/AB = AE/AC = 2/5
∠A = ∠A
[Common]
ΔADE∼ΔABC (By SAS similarity)
AD/AB = DE/BC
[Since, corresponding sides of two similar triangles are proportional]
8/20 = DE/BC
⅖ = DE/BC
2BC = 5DE
BC = 5/2DE
HOPE THIS ANSWER WILL HELP YOU...
Attachments:
Answered by
7
As triangle ADE ~ ABC
AD/AB = DE/BC
8/8+12 = DE/BC
8/20 = DE/BC
2/5 = DE/BC
BC = 5/2 DE
AD = 6 cm; BD = 9 cm
AE = 8 cm and CE = 12 cm
Now we have:
AD/BD = 6/9 = 2/3
AE/CE = 8/12 = 2/3
So AD/BD = AE/CE
We know that if a line intersects any two sides of a triangle in equal ratio then the line is parallel to the third side.
DE = BC
Similar questions