Question

In $\Delta ABC$, $\angle ABC = 135^\circ $,
Prove that $A{C^2} = A{B^2} + B{C^2} + 4\left( {\Delta ABC} \right)$.

Answer 1

Hint:
 Draw a triangle having $\angle ABC = 135^\circ $ and name the triangle as $ABC$, extend the side $BC$ and draw a perpendicular on it from point $A$ and use Pythagoras theorem on triangle $ADB$ and $ADC$
Complete step-by-step answer:
Given:\[\Delta ABC\] and \[\angle ABC = 135^\circ \]
To prove:\[{(AC)^2} = {(AB)^2} + {(BC)^2} + 4ar(\Delta ABC)\]
Construction: Draw a triangle $ABC$ and produce BC to D and draw AD perpendicular to DC.

Proof: Consider the triangle $ADB$ and apply Pythagoras theorem on \[\Delta ADB\].
\[{(AB)^2} = {(AD)^2} + {(DB)^2}\]
Now, consider the triangle $ADC$ and again, applying Pythagoras theorem on \[\Delta ADC\].
\[{(AC)^2} = {(AD)^2} + {(DC)^2}\] (1)
As, side $DC$ is equal to the sum of sides $DB$ and $BC$, that is, \[DC = DB + BC\].
So, simplify the obtained equation (2) by substituting \[DC = DB + BC\].
\[{(AC)^2} = {(AD)^2} + {(DB + BC)^2}\]
Now, apply the algebraic identity \[{(a + b)^2} = {a^2} + {b^2} + 2a.b\] and simplify the obtained equation further.
\[{(AC)^2} = {(AD)^2} + {(DB)^2} + {(BC)^2} + 2 \cdot DB \cdot BC\]
It is observed that \[\angle ADB = 90^\circ \], because $AD$ is perpendicular to $BC$ or $DB$.
Now, find the angle $\angle ABD$.
As given $\angle ABC = 135^\circ $.
$\angle DBC$ is a straight line angle, which is equal to $180^\circ $.
So sum of the angles $\angle ABD$ and $\angle ABC$ will be equal to $180^\circ $.
That is $\angle ABD + \angle ABC = 180^\circ $
Now, find $\angle ABD$ by using $\angle ABC = 135^\circ $.
$
  \angle ABD + 135^\circ = 180^\circ \\
   \Rightarrow \angle ABD = 180^\circ - 135^\circ \\
   \Rightarrow \angle ABD = 45^\circ \\
 $
Now find the angle $DAB$ with the help of the obtained angle and \[\angle ADB = 90^\circ \].
as the sum of all angles of a triangle is $180^\circ $.
Then,
\[\angle ADB + \angle DAB + \angle ABD = 180^\circ \]
Now simplify the obtained equation for angle $DAB$,
\[
  90^\circ + \angle DAB + 45^\circ = 180^\circ \\
  135^\circ + \angle DAB = 180^\circ \\
  \angle DAB = 180^\circ - 135^\circ \\
   = 45^\circ \\
\]

As a result, \[ \Rightarrow \angle ABD = \angle DAB = 45^\circ \]
This implies, \[AD = DB\](Sides opposite to equal angles are equal) (2)
From equation (1) and (2)
\[
  {(AC)^2} = {\left( {AD} \right)^2} + {\left( {DC} \right)^2} \\
  {(AC)^2} = {\left( {AD} \right)^2} + {\left( {DB + BC} \right)^2}{\text{ As}}\left( {DC = DB + BC} \right) \\
  {(AC)^2} = {(AD)^2} + {(DB)^2} + {\left( {BC} \right)^2} + 2 \cdot DB \cdot BC{\text{ As}}\left[ {{{\left( {DB + BC} \right)}^2} = D{B^2} + B{C^2} + 2.DB.BC} \right]{\text{ }} \\
  {\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2} + 2.AD.BC{\text{ As}}\left( {A{C^2} = A{B^2} + B{C^2}} \right){\text{ and }}\left( {AD = DB} \right) \\
\]
Multiply and divide by 2 in the term \[2 \cdot AD \cdot BC\] of the obtained expression.
\[
  {(AC)^2} = {(AB)^2} + {(BC)^2} + \dfrac{2}{2}(2 \cdot AD \cdot BC) \\
  {(AC)^2} = {(AB)^2} + {(BC)^2} + 4\left( {\dfrac{1}{2} \cdot AD \cdot BC} \right) \\
  {(AC)^2} = {(AB)^2} + {(BC)^2} + 4ar\left( {\Delta ABC} \right) \\
\]
Hence proved that \[{(AC)^2} = {(AB)^2} + {(BC)^2} + 4ar(\Delta ABC)\].

Note:
Use the Pythagoras Theorem\[{(hypotenuse)^2} = {(base)^2} + {(perpendicular)^2}\] and use the fact that Theorem based on opposite sides triangles, also use the formula of area of area of triangle.