Question

The vertices of a triangle ABC are (3, 0) , (0, 6) and (6, 9) . The line DE divides AB and AC in the ratio of 1 : 2.Prove that:
∆ABC = 9∆ADE ​

Answer 1

Question :

The vertices of a triangle ABC are (3, 0) , (0, 6) and (6, 9) . The line DE divides AB and AC in the ratio of 1 : 2.Prove that:

∆ABC = 9∆ADE ​

Given :

• Vertices of triangle ABC are (3,0) , (0,6) and (6,9)
• The line DE divides AB and AC in the ratio 1:2

To prove :

• ΔABC = 9ΔADE

Solution :

As DE divides AB and AC in the ratio of 1 : 2, so m₁ = 1 and m₂ = 2

Line DE divides AB and AC in the ratio 1 : 2 , so vertices of D :

$: \implies\sf \bigg(\dfrac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\bigg) \ , \ \bigg(\dfrac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\bigg) \\\\ \\ : \implies\sf \bigg(\dfrac{1 \times 0 + 2 \times 3}{1 + 2} \bigg) \ , \ \bigg(\dfrac{1 \times 6 + 2 \times 0}{1 + 2} \bigg) \\\\ \\ : \implies\sf \bigg(\dfrac{6}{3} \bigg)\ , \ \bigg(\dfrac{6}{3}\bigg) \\\\ \\ : \implies\large\underline{\boxed{\sf{2,2}}} \\\\ \\ \therefore\underline{\sf{The \ vertices \ of \ D = (2,2) }}$

Line DE divides AB and AC in the ratio of 1 : 2 , so vertices of E :

$: \implies\sf \bigg(\dfrac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\bigg) \ , \ \bigg(\dfrac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \bigg) \\\\ \\ : \implies\sf \bigg(\dfrac{1 \times 6 + 2 \times 3}{1 + 2} \bigg) \ , \ \bigg(\dfrac{1 \times 9 + 2 \times 0}{1 + 2}\bigg) \\\\\ \\ :\implies\sf \bigg(\dfrac{12}{3}\bigg) \ , \ \bigg(\dfrac{9}{3}\bigg) \\\\ \\ : \implies\large\underline{\boxed{\sf{4,3 }}} \\\\\\ \therefore\underline{\sf{The \ vertices \ of \ E = (4,3) }}}$

Now in ΔABC,

∴ x₁ = 3 , x₂ = 0, x₃ = 6

∴ y₁ = 0, y₂ = 6, y₃ = 9

And in ΔADE,

∴ x₁ = 3, x₂ = 2, x₃ = 4

∴ y₁ = 0, y₂ = 2, y₃ = 3

Now area of ΔABC :-

$: \implies\displaystyle\sf \dfrac{1}{2} |x1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3| \\\\ \\ : \implies\sf \dfrac{1}{2} |3 \times 6 + 0 \times 9 + 6 \times 0 - 0 \times 0 - 6 \times 6 - 3 \times 9| \\\\ \\ : \implies\sf \dfrac{1}{2} |18 - 36 - 27| \\\\ \\ : \implies\sf \dfrac{1}{2} |-45| \\\\ \\ : \implies\large\underline{\boxed{\sf{\dfrac{45}{2} \ unit^2 }}}$

Now area of ΔADE :

$: \implies\sf \dfrac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \\\\ \\ : \implies\sf \dfrac{1}{2} |3(2 - 3) + 2(3 - 0) + 4(0 - 2)| \\\\ \\ : \implies\sf \dfrac{1}{2} |3 \times (-1) + 2 \times 3 + 4 \times (-2)| \\\\ \\ : \implies\sf \dfrac{1}{2} |-3 + 6 - 8| \\\\ \\ : \implies\sf \dfrac{1}{2} |5| \\\\ \\ : \implies\large\underline{\boxed{\sf{\dfrac{5}{2} \ unit^2 }}}$

Now comparing areas of ΔABC and ΔADE :

$: \implies\sf 9 \times \dfrac{5}{2} = \dfrac{45}{2} \\\\ \\ : \implies\sf 9 \times \triangle ADE = \triangle ABC \\\\ \\ : \implies\large\underline{\boxed{\bold{\triangle ABC = 9 \triangle ADE }}} \qquad[\sf Hence \ proved]$