13 In Figure P13 prove that h2 cc2 identity 7 above C2 D c F

13. In Figure P.13 prove that h2 cc2 (identity (7) above) C2 D c Figure P.13

Solution

We use Pythagoras\'s theorem for this sum. Base² + Height² = Hypotenuse²

In Triangle ABC : a² + b² = c². But c = c₁ + c₂. Therefore a² + b² = (c₁+c₂)² = c₁² + c₂² + 2c₁c₂-----(1)

In Triangle CBD: h² + c₁² = a². Therefore h² = a² - c₁².------(2)

In Triangle CAD: h² + c₂² = b². Therefore h² = b² - c₂².------(3)

Adding (2) + (3) (both left hand side and right hand side)

2*h² = a² - c₁² + b² - c₂² = a² + b² - c₁² - c₂². Substituting for a² + b² from (1), we get

h² + *h² = c₁² + c₂² + 2c₁c₂ - c₁² - c₂² = 2c₁c₂

So 2 * h² = 2 * c₁c₂

Therefore h² = c₁ * c₂