Ex.6.5 Q14 Triangles Solution - NCERT Maths Class 10

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Question

The perpendicular from \(A\) on side \(BC\) of a \(\Delta ABC\) intersects \(BC\) at \(D\) such that \(DB\) \(=\) \(3CD\)(see Fig. below). Prove that \(2A{{B}^{2}}=2A{{C}^{2}}+B{{C}^{2}}\) .

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 \(\Delta ABC\,\,,\,\,AD\bot BC\) and \(\,BD=3CD\)

\[\begin{align}BD + CD &= BC\\3CD + CD &= BC\\4CD &= BC\\CD &= \frac{1}{4}BC \ldots  \ldots  \ldots .\left( 1 \right)\\{\rm{and,\qquad }}BD &= \frac{3}{4}BC \ldots  \ldots  \ldots .\left( 2 \right)\end{align}\]

In \(\Delta ADC\),

\[\begin{array}{l}A{C^2} = AD{}^2 + C{D^2}{\rm{      }}\left[ {\angle ADC = {{90}^0}} \right]\\A{D^2} = A{C^2} - C{D^2}{\rm{      }} \ldots  \ldots  \ldots .\left( 3 \right)\end{array}\]

In \(\Delta ADB\),

\[\begin{align}A{B^2} &= AD{}^2 + B{D^2}{\rm{                              }}\qquad\qquad\qquad\qquad\left[ {\angle ADB = {{90}^0}} \right]\\A{B^2} &= A{C^2} - C{D^2} + B{D^2}{\rm{                   }}\qquad\qquad\qquad\left[ {{\rm{from (3)}}} \right]\\\,\,\,\,\,\,\,\,\,A{B^2}\,\, &= A{C^2} + {\left( {\frac{3}{4}BC} \right)^2} - {\left( {\frac{1}{4}BC} \right)^2}{\rm{     }}\qquad\left[ {{\rm{from (1) and (2)}}}\right]\\
A{B^2} &= A{C^2} + \frac{{9B{C^2} - B{C^2}}}{{16}}\\A{B^2} &= A{C^2} + \frac{{8B{C^2}}}{{16}}\\A{B^2} &= A{C^2} + \frac{1}{2}B{C^2}\\2A{B^2} &= 2A{C^2} + B{C^2}\end{align}\]