Question

In right triangle ABC right angled at B , points P&Q trisect BC. then 3AC² + 5AP² =

Answer 1

Given right DABC, right angled at D and E are points of trisection of the side BC
Let BD = DE = EC = k
Hence we get BE = 2k and BC = 3k
In ΔABD, by Pythagoras theorem, we get
AD2�= AB2�+ BD2
AD2�= AB2�+ k2
Similarly, in ΔABE we get
AE2�= AB2�+ BE2
Hence AE2�= AB2�+ (2k)2
= AB2�+ 4k2�and
AC2�= AB2�+ BC2
�= AB + (3k)2
AC2= AB2�+ 9k2
Consider, 3AC2�+ 5AD2�= 3(AB2�+ 9k2) + 5(AB2�+ 4k2)
= 8AB2�+ 32k2
= 8(AB2�+ 4k2)
∴ 3AC2�+ 5AD2�= 8AE2
[Thank You]