Question

In the given figure, ABC is a triangle angled at B.D and E are ponts on BC trisect it.
Prove that $8{AE}^2=3{AC}^2+5{AD}^2.$

Answer 1

Given right DABC, right angled at B

D & E are points of trisection of the side BC.

Let BD$=$DE$=$EC$=$k

Hence we get BE$=2k$ & BC$=3k$

In $\Delta$ABD by Pythagoras theorem we get

$A{D}^2=A{B}^2+B{D}^2$

$A{D}^2=A{B}^2+k^2$

Similarly, in $\Delta$ABE we get

$A{E}^2=A{B}^2+B{E}^2$

Hence $A{E}^2=A{B}^2+(2k{)}^2$

$=A{B}^2+4k^2$ and

$A{C}^2=A{B}^2+B{C}^2$

$=AB+(3k{)}^2$

$A{C}^2=A{B}^2+9k^2$

Consider $3A{C}^2+5A{D}^2=3(A{B}^2+9k)+5(A{B}^2+4k^2)$

$=8A{B}^2+32k^2$

$=8(A{B}^2+4k^2)$

$\therefore 3A{C}^2+5A{D}^2=8A{E}^2$.