Ex. 6.6 Q5 Triangles Solution - NCERT Maths Class 10

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Question

In , \(AD\) is a median of a triangle \(ABC\) and \(AM \bot BC.\)

Prove that:

\(\begin{align} i)\;\;\;\;A{C^2} &= A{D^2} + BC.DM + {\left( {\frac{{BC}}{2}} \right)^2}\\ ii)\;\;\;A{B^2} &= A{D^2} - BC.DM + {\left( {\frac{{BC}}{2}} \right)^2}\\ iii)\;\;A{C^2} &+ A{B^2} = 2A{D^2} + \frac{1}{2}B{C^2} \end{align}\)

 

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:

1. In \(\Delta AMC\)

\[\begin{align} \angle AMC &= {90^ \circ }\\ A{C^2} &= A{M^2} + M{C^2}\\ &= A{M^2} + {\left[ {MD + CD} \right]^2}\\ &= A{M^2} + M{D^2} + C{D^2} + 2MD.CD\\ &= A{D^2} + {\left( {\frac{{BC}}{2}} \right)^2} + 2MD.\left( {\frac{{BC}}{2}} \right) \end{align}\]

Since, In \(\Delta AMD,{\rm{ }}A{D^2} = A{M^2} + D{M^2}\), and  is the midpoint of  means \(\begin{align}\,\,BD = CD = \frac{{BC}}{2} \\ A{C^2} = A{D^2} + MD.BC + {\left( {\frac{{BC}}{2}} \right)^2} .....(i) \end{align} \)

2. In \(\Delta \,AMB\)

\[\begin{align}\angle AMB &= {90^ \circ }\\A{B^2} &= A{M^2} + B{M^2}\\ &= A{M^2} + {\left[ {BD - DM} \right]^2}\\ &= A{M^2} + B{D^2} + D{M^2} - 2BD.DM\\ &= A{M^2} + D{M^2} + {\left( {\frac{{BC}}{2}} \right)^2} - 2 { \left({ \frac{{BC}}{2} }\right)} DM \end{align}\]

Since,In \(\Delta AMD, AD^2 = AM^2 + DM^2 \;\text{and}\;D\) is the midpoint of \(BC\) means \(BD = CD = \frac{{BC}}{2}\)

\[\begin{align}A{B^2} = A{D^2} - BC.DM + {\left( {\frac{{BC}}{2}} \right)^2} ....\left(\rm {ii} \right)\end{align}\]

3. Adding (i) and (ii)\[\begin{align} A{C^2} + A{B^2} &= A{D^2} + {\left( {\frac{{BC}}{2}} \right)^2} + BC.DM + A{D^2} + {\left( {\frac{{BC}}{2}} \right)^2} - BC.DM\\ A{C^2} + A{B^2} &= 2A{D^2} + 2{\left( {\frac{{BC}}{2}} \right)^2}\\ A{C^2} + A{B^2} &= 2A{D^2} + \frac{1}{2}{B{C^2}} \end{align}\]

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