ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:

1) ΔADE ~ ΔACB

2) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

3) Find. Area of ΔADE: area of quadrilateral BCED

#### Solution

1) Consider ΔADE and ΔACB

∠A = ∠A [Common]

m∠A = m∠E = 90°

Thus by Angle-Angle similarity, triangles ΔACB ~ ΔADE

2) Since ΔADE ~ ΔACB, their sides are proportional

`=> (AE)/(AB) = (AD)/(AC) = (DE)/(BC)` ....(1)

In ΔABC, by Pythagoras Theorem, we have

`AB^2 + BC^2 = AC^2`

`=> AB^2 + 5^2 = 13^2`

`=> AB = 12 cm`

From equation 1 we have

`4/12 = (AD)/13 = (DE)/5`

`=> 1/3 = (AD)/13`

`=> AD = 13/3 cm`

Also `4/12 = (DE)/5`

`=> DE = 20/12 = 5/3 cm`

3) We need to find the area of ADE and quadrilateral BCED

Area of ΔADE = `1/2 xx AE XX DE = 1/2 xx 4 xx 5/3 = 10/3 cm^3`

Area of quad.BCED = Area of ΔABC Area of ΔADE

` = 1/2 xx BC xx AB - 10/3`

`= 1/2 xx 5 xx 12 - 10/3`

`= 30 - 10/3`

`= 80/3 cm^2`

Thus ratio of areas of ADE to quadrilateral BCED = `(10/3)/(80/3) = 1/8`