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 \(\,BD=3CD\)

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

 

In \(\Delta ADC\),

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

 

In \(\Delta ADB\),

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