Ex. 6.6 Q3 Triangles Solution - NCERT Maths Class 10

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Question

In Fig. below, \(ABC\) is a triangle in which \(\Delta ABC > 90°\) and \(AD \perp CB\) produced.

 Prove that:

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

Text Solution

Reasoning:

Pythagoras theorem

Steps:

In \(\Delta ADC\)

\[\begin{align} \angle ADC &= {90^ \circ }\\ \Rightarrow A{C^2} &= A{D^2} + C{D^2}\\ &= A{D^2} + {\left[ {BD + BC} \right]^2}\\ &= A{D^2} + B{D^2} + B{C^2} + 2BC \cdot BD\\ A{C^2} &= A{B^2} + B{C^2} + 2BC \cdot BD \end{align}\]

\(\begin{align}(\therefore \,{\text{In}}\; \angle ADB,\;A{B^2} = A{D^2} + B{D^2}\;\end{align}\) \()\)

  
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