Question
In the given figure, $\angle\text{B}<90^\circ$ and segment $\text{AD}\perp\text{BC}$ show that
- $b^2 = h^2 + a^2 + x^2 - 2ax$
- $b^2 = a^2 + c^2 - 2ax$
✓
Answer
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,
$b^2 = h^2 + (a - x)^2$
$b^2 = h^2 + a^2 - 2ax + x^2$
$b^2 = h^2 + a^2 + x^2 - 2ax .....(2)$
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, $b^2 = c^2 + a^2 - 2ax.$
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,
$b^2 = h^2 + (a - x)^2$
$b^2 = h^2 + a^2 - 2ax + x^2$
$b^2 = h^2 + a^2 + x^2 - 2ax .....(2)$
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, $b^2 = c^2 + a^2 - 2ax.$
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