Ex.6.5 Q13 Triangles Solution - NCERT Maths Class 10

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



Text Solution


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


In \(\Delta ABC,\,\,\,\,\,\angle ABC={{90}^{0}}\)

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

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

In \(\Delta ACE\) ,

\(AE{}^{2}=A{{C}^{2}}+C{{E}^{2}}\) (Pythagoras theorem)……….(1)

In \(\Delta DCB\)

\(BD{}^{2}=C{{D}^{2}}+B{{C}^{2}}\) ……………..(2)

Adding (1) and (2)

\(\begin{align} & AE{}^{2}+B{{D}^{2}}=A{{C}^{2}}+CE{}^{2}+C{{D}^{2}}+B{{C}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=A{{C}^{2}}+BC{}^{2}+E{{C}^{2}}+C{{D}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=A{{B}^{2}}+DE{}^{2} \\ \end{align}\)

\(\begin{align} & [\text{In}\,\,\Delta ABC,\,\,\,\,\,\,\,\,\,\,\,\angle C={{90}^{0}}\Rightarrow A{{C}^{2}}+BC{}^{2}=A{{B}^{2}}\,\text{and} \\ & \text{In}\,\,\Delta CDE,\,\,\angle DCE={{90}^{0}}\Rightarrow C{{D}^{2}}+C{{E}^{2}}=D{{E}^{2}}] \\ \end{align}\)

\(\begin{align} \Rightarrow A{{E}^{2}}+BD{}^{2}=A{{B}^{2}}+D{{E}^{2}}\end{align}\)