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`
