Question

In a right angled triangle ABC , right angled at B , points D & E divides BC BA respectively in the ratio 2:1 . Prove that 9 AD^2 + 9CE^2= 13 AC^2

Answer 1

Question:

In a right-angled triangle ABC, right-angled at B, points D & E divides BC & BA respectively in the ratio 2:1. Prove that 9AD²+ 9CE² = 13AC²

Solution:

(Figure attached for reference)

Given:

In a ΔABC,

∠B = 90°

BD : DC = 2 : 1

BE : EA = 2 : 1

To Prove:

9AD²+ 9CE² = 13AC²

Proof:

ATQ, D & E divides BC & BA respectively in the ratio 2:1, Therefore,

$\Longrightarrow \sf \ \dfrac{DC}{BD} = \dfrac{1}{2}\\ \\ \\\sf Add \ 1 \ on \ both \ sides.\\ \\ \\\Longrightarrow \sf \ \dfrac{DC}{BD} + 1 = \dfrac{1}{2} + 1\\ \\ \\\Longrightarrow \sf \ \dfrac{DC + BD}{BD} = \dfrac{1 + 2}{2}\\ \\ \\\Longrightarrow \sf \ \dfrac{BC}{BD} = \dfrac{3}{2}\\ \\ \\\Longrightarrow \sf \ BD = \dfrac{2}{3} \ BC$

Similarly,

$\Longrightarrow \sf \ EB = \dfrac{2}{3} \ AB$

In ΔABD

∠B = 90°

By using Pythagoras Theorem we get;

$\Longrightarrow \sf AD^2 = AB^2 + BD^2\\ \\ \\\Longrightarrow \sf AD^2 = AB^2 + \bigg{\{} \dfrac{2}{3} \ BC \bigg{\}}^2\\ \\ \\\Longrightarrow \sf AD^2 = AB^2 + \dfrac{4}{9} \ BC^2\\ \\ \\\Longrightarrow \sf AD^2 = \dfrac{9AB^2 + 4BC^2}{9}\\ \\ \\\Longrightarrow \sf 9AD^2 = 9AB^2 + 4BC^2 \longmapsto \textcircled{\scriptsize \sf 1}$

In ΔEBC

∠B = 90°

By using Pythagoras Theorem we get;

$\Longrightarrow \sf EC^2 = EB^2 + BC^2\\ \\ \\\Longrightarrow \sf EC^2 = \bigg{\{} \dfrac{2}{3} \ AB \bigg{\}}^2 + BC^2\\ \\ \\\Longrightarrow \sf EC^2 = \dfrac{4}{9} \ AB^2 + BC^2\\ \\ \\\Longrightarrow \sf EC^2 = \dfrac{4AB^2 + 9BC^2}{9}\\ \\ \\\Longrightarrow \sf 9EC^2 = 4AB^2 + 9BC^2 \longmapsto \textcircled{\scriptsize \sf 2}$

Adding equation 1 and 2 we get,

⇒ 9AD² + 9EC² = 9AB² + 4BC² + 4AB² + 9BC²

⇒ 9AD² + 9EC² = 13AB² + 13BC²

⇒ 9AD² + 9EC² = 13(AB² + BC²)

ATQ to Pythagoras Theorem, AB² + BC² = AC²

⇒ 9AD² + 9EC² = 13AC²

Hence Proved!

(Refer to the attachment for markings)

Hope you understood! (◕‿◕)