Ex.6.5 Q13 Triangles Solution - NCERT Maths Class 10

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Question

\(D\) and \(E\) are points on the sides \(CA\) and \(CB\) respectively of a triangle \(ABC\) right angled at \(C\). Prove that  \(\begin{align} A E^{2}+B D^{2}=A B^{2}+D E^{2}\end{align}\).

Diagram

Text Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

In \(\Delta ABC,\,\,\angle ABC={{90}^{\circ}}\)

\(D\), \(E\) are points on \(AC\) and \(BC\)

Join \(AE\), \(DE\) and \(BD\)

 

In \(\Delta ACE\) ,

\[\begin{align} AE{}^{2} & =A{{C}^{2}}+C{{E}^{2}} \;\;\dots (1) \\ & [ \text { Pythagoras } \text{ theorem } ] \end{align} \]

 

In \(\Delta DCB\)

\[\begin{align} BD{}^{2}=C{{D}^{2}}+B{{C}^{2}} \;\;\dots (2) \end{align} \]

 

Adding \((1)\) and \((2)\)

\[\begin{align} & AE{}^{2}+B{{D}^{2}} \\ & \;\;= \! A{C}^{2}\!+ \! CE{}^{2} \! +\! C{D}^{2} \!\!+ \! B{{C}^{2}}  \\ & \;\;= \! A{C}^{2} \! + \! BC{}^{2} \! + \! E{{C}^{2}} \! + \! C{{D}^{2}} \\ & \;\;=A{{B}^{2}}+DE{}^{2} \\ \end{align}\]

\[\begin{align} & \text{In}\,\,\Delta ABC,\,\,\,\,\, \angle C={{90}^{\circ}} \\ & \Rightarrow A{{C}^{2}}+BC{}^{2}=A{{B}^{2}} \\\\ & \qquad \qquad  \text{and} \\\\ & \text{In}\,\,\Delta CDE,\,\,\angle DCE={{90}^{\circ}} \\ & \Rightarrow C{{D}^{2}}+C{{E}^{2}}=D{{E}^{2}}\\ &\Rightarrow A{{E}^{2}}+BD{}^{2}=A{{B}^{2}}+D{{E}^{2}}\  \end{align}\]