Ex. 6.6 Q4 Triangles Solution - NCERT Maths Class 10

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Question

In Fig below, \(ABC\) is a triangle in which \(ABC < 90°\) and \(AD \bot BC.\)
Prove that:

\(A{C^2} = A{B^2} + B{C^2} - 2BC \cdot BD\)

Text Solution

 

Reasoning:

Pythagoras Theorem.

Steps:

In \(\Delta ADC\)

\[\begin{align} \angle ADC &= {90^ \circ }\\ A{C^2}& = A{D^2} + D{C^2}\\ &= A{D^2} + {\left[ {BD - BC} \right]^2}\\ &= A{D^2} + B{D^2} + B{C^2} - 2BC.BD\\ A{C^2}&= A{B^2} + B{C^2} - 2BC.BD \;\;\;\;\;{\rm{ }}\left[ {{\rm{In }}\;\Delta ADB,\,\,\,A{B^2} = A{D^2} + B{D^2}} \right]\end{align}\] \(\)

  
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