Question

In the given figure, ∠B < 90° and segment AD ⊥ BC, show that

(i) b² = h² + a² + x² - 2ax

(ii) b² = a² + c² - 2ax

Answer 1

(i) Since AD perpendicular to BC we obtained two right angled triangles, triangle ADB and triangle ADC.

We will use Pythagoras theorem in the right angled triangle ADC

AC² = AD² + DC²              ......(1)

Let us substitute AD = h, AC = b and DC = (a − x) in equation (1) we get,

b² = h² + (a − x)²

b² = h² + a² − 2ax + x²

b² = h² + a² + 𝑥² − 2ax                ......(2)

(ii) Let us use Pythagoras theorem in the right angled triangle ADB as shown below,

AB² = AD² + BD²                       ......(3)

Let us substitute AB = c, AD = h and BD = x in equation (3) we get,

c² = h² + x²

Let us rewrite the equation (2) as below,

b² = h² + x² + a² - 2ax                ......(4)

Now we will substitute h² + x² = c² in equation (4) we get,

b² = c² + a² - 2ax

Therefore, b² = c² + a² - 2ax.