In the given figure, <90^ and segment AD show that 1. b^2 = h^2 + a^2 + x^2 - 2ax 2. b^2 = a^2 + c^2 - 2ax

1. 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) 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.

In the given figure, <90^ and segment AD show that 1. b^2 = h^2 +…

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Question

In the given figure, $\angle\text{B}<90^\circ$ and segment $\text{AD}\perp\text{BC}$ show that

$b^2 = h^2 + a^2 + x^2 - 2ax$
$b^2 = a^2 + c^2 - 2ax$

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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, $b^2 = c^2 + a^2 - 2ax.$