#### Question

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

(i) b^{2 }= h^{2 }+ a^{2 }+ x^{2 }- 2ax

(ii) b^{2} = a^{2} + c^{2} - 2ax

#### Solution

(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

AC^{2} = AD^{2} + DC^{2} ......(1)

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

b^{2} = h^{2} + (a − x)^{2}

b^{2} = h^{2} + a^{2} − 2ax + x^{2}

b^{2} = h^{2} + a^{2} + 𝑥^{2} − 2ax ......(2)

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

AB^{2} = AD^{2} + BD^{2} ......(3)

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

c^{2} = h^{2} + x^{2}

Let us rewrite the equation (2) as below,

b^{2} = h^{2} + x^{2} + a^{2} - 2ax ......(4)

Now we will substitute h^{2} + x^{2} = c^{2} in equation (4) we get,

b^{2} = c^{2} + a^{2} - 2ax

Therefore, b^{2} = c^{2} + a^{2} - 2ax.