In ΔABC, it is given that AB = 12 cm, ∠B = 90° and AC = 15 cm. If D and E are points on AB and AC respectively such that ∠AED = 90° and DE = 3 cm, prove that :

(i) ΔABC ∼ ΔAED.

(ii) ar(ΔAED) = 6 cm².

(iii) ar(quad BCED) : ar(ΔABC) = 8 : 9.

(i) Given,

∠ABC = ∠AED = 90° [Given]

∠BAC = ∠DAE [Common angle]

∴ ΔABC ∼ ΔAED (By A.A. axiom)

Hence, proved that ΔABC ∼ ΔAED.

(ii) ΔABC is right-angled triangle, applying pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ BC² = AC² - AB²

⇒ BC² = (15)² - (12)²

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = $\sqrt{81}$

⇒ BC = 9 cm

Area of ΔABC = $\dfrac{1}{2}$ × Base × height

= $\dfrac{1}{2}$ × 12 × 9

= 54 cm²

Since, ΔABC ∼ ΔAED,

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\therefore \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{{ED}^2}{{BC}^2} \\[1em] \Rightarrow \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{3^2}{9^2} \\[1em] \Rightarrow \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{9}{81} \\[1em] \Rightarrow \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{1}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = \text{ar(ΔABC)} \times \dfrac{1}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = \dfrac{54}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = 6 \text{ cm}^2$

Hence, proved that ar(ΔAED) = 6 cm².

(iii) From figure,

ar(quad. BCED) = ar(ΔABC) - ar(ΔAED)

ar(quad. BCED) = 54 - 6

ar(quad. BCED) = 48 cm².

$\Rightarrow \dfrac{\text{ar(quad. BCED)}}{\text{ar.(ΔABC)}} = \dfrac{48}{54} = \dfrac{8}{9}$.

Hence, proved that ar(quad BCED) : ar(ΔABC) = 8 : 9.