Question

In the given figure , D and E trisect BC . Prove that 8AE^2 = 3AC^2+5AD^2​

Answer 1

GiveN :-

• In the given figure , D and E trisect BC .

To ProvE :-

• 8AE² = 3AC² + 5AD²

ProoF :-

Given that , In the given figure , D and E trisect BC . We need to prove that 8AE² = 3AC² + 5AD² . It's pretty clear that we have to make use of the Pyhtagoras Theorem or Baudhayan's Theorem . This Theorem states that in a right angled triangle the sum of squares of base and perpendicular is equal to the square of hypontenuse. That is ,

$\sf\pink{\dashrightarrow Hypontenuse^2 = Perpendicular^2+Base^2 }$

$\red{\frak{In \ figure }}\begin{cases}\sf AC^2= AB^2+BC^2 \\\sf AE^2=AB^2+BE^2\\\sf AD^2= AB^2+BD^2 \end{cases}$

• D I A G R A M :-

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Let's start by taking RHS and try to get LHS :-

$\red{\bigstar}\underline{\sf Taking \ RHS }$

$\sf\to 3AC^2 + 5AD^2 \\\\\to\sf 3( AB^2+BC^2) + 5(AB^2 + BD^2 ) \\\\\to \sf 3AB^2 + 3BC^2+5AB^2+5BD^2 \\\\\sf \to 8AB^2 + 3BC^2 + 5BD^2$

So here we got 8AB² + 3BC² + 5BD². Now basically we need to convert 3BC² + 5BD² into 8BE² because 8AB² + 8BE² = 8AE² .

$\sf\to 8AB^2 + 3\bigg( \dfrac{3BE}{2}\bigg)^2 + 5\bigg(\dfrac{BE}{2}\bigg)^2 \\\\\to \sf 8AB^2 + 3\times \dfrac{9BE^2}{4}+\dfrac{5BE^2}{4} \\\\\to\sf 8AB^2 + \dfrac{27BE^2}{4}+\dfrac{5BE^2}{4}\\\\\to\sf 8AB^2 + \dfrac{ 27BE^2+5BE^2}{4}\\\\\to\sf 8AB^2 +\dfrac{32BE^2}{4} \\\\\sf\to 8AB^2 + 8BE^2 \\\\\to \sf 8( AB^2+BE^2) \\\\\to \sf \pink{8AE^2 }$

$\sf\qquad\qquad\red{\bigstar} \:\: = RHS$

$\boxed{\green{\bigstar}\textsf{\textbf{\orange{ Hence Proved !! }}}</p><p>\green{\bigstar}}$