Question

In the figure $\angle B={90}^{\circ }$, D and E trisect BC, Prove that $8{AE}^2=3{AC}^2+5{AD}^2$ [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Proof : 2 Marks

Since D and E are the points of trisection of BC. Therefore,

BD = DE = CE.

Let BD = DE = CE = x.

Then, BE = 2x and BC = 3x.

In right triangles ABD, ABE and ABC, we have

$\Rightarrow {AD}^2={AB}^2+{BD}^2$

$\Rightarrow {AD}^2={AB}^2+x^2\dots \dots \left(1\right)$

${AE}^2={AB}^2+{BE}^2$

$\Rightarrow {AE}^2={AB}^2+4X^2\dots \dots \left(2\right)$

and, ${AC}^2={AB}^2+{BC}^2$

$\Rightarrow {AC}^2={AB}^2+9X^2\dots \dots \left(3\right)$

Now consider,

$8{AE}^2-3{AC}^2-5{AD}^2$

$=8({AB}^2+4x^2)-3({AB}^2+9x^2)-5({AB}^2+x^2)$

$=8{AB}^2+32x^2-3{AB}^2-27x^2-5{AB}^2-5x^2$

$=8{AB}^2+32x^2-8{AB}^2-32x^2=0$

$\Rightarrow 8{AE}^2-3{AC}^2-5{AD}^2=0$

$\Rightarrow 8{AE}^2=3{AC}^2+5{AD}^2$