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:

i) \(\begin{align} A{C^2} = A{D^2} + BC.DM + {\left[ {\frac{{BC}}{2}} \right]^2} \end{align}\)

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

iii) \(\begin{align} A{C^2} + A{B^2} = 2A{D^2} + \frac{1}{2}B{C^2} \end{align}\)

 Video Solution
Triangles
Ex 6.6 | Question 5

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:

\(\rm i)\) In \(\Delta AMC\)

\(\angle AMC = {90^ \circ }\)

\[\begin{align} A{C^ 2} &\!=\!A{M^2}\!+\! M{C^2}\\ &\!= \!\! A{M^2} \!+\! {\left[ {MD \!+\! CD} \right]^2} \\ &\!=\! \!A{M^2} \!+\! M{D^2} \!+\! C{D^2} +\! 2MD.CD \\ &\!=\! \!A{D^2} \!+\! {\left[ {\frac{{BC}}{2}} \right]^2} \! +\! 2MD.\left[ {\frac{{BC}}{2}} \right]\end{align}\]

Since, in \(\Delta AMD,{\rm{ }}A{D^2} = A{M^2} + D{M^2}\), and  is the midpoint of  means

\(\begin{align}BD = CD = \frac{{BC}}{2} \end{align} \)

\[\begin{align} A{C^2}\! = \! A{D^2} \!+\! MD.BC \!+\! {\left[ {\frac{{BC}}{2}} \right]^2} \;\dots\rm (i) \end{align}\]

\(\rm ii)\) In \(\Delta \,AMB\)

\(\angle AMB = {90^ \circ }\)

\[\begin{align}A{B^2} &\!=\!A{M^2}\! +\! B{M^2}\\ &\!=\! A{M^2}\! +\! {\left[ {BD \!-\! DM} \right]^2}\\ &\!=\! A{M^2}\! +\! B{D^2}\! +\! D{M^2} \! - \!2BD.DM \\ &\!= \! \begin{Bmatrix}A{M^2}\! +\! D{M^2} \!+\! {\left[ {\frac{{BC}}{2}} \right]^2} \!\\-\! 2 { \left[{ \frac{{BC}}{2} }\right]} DM \end{Bmatrix}  \end{align}\]

Since,

In \(\Delta AMD,\)

 \( AD^2 = AM^2 + DM^2 \text{and}\;D\) 

is the midpoint of \(BC\) means

 \(BD = CD = \frac{{BC}}{2}\)

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

\(\rm iii)\) Adding (\(\rm{i}\)) and (\(\rm{ii}\))

\[\begin{align} A{C^2}\!+\!A{B^2} &\!=\!\begin{Bmatrix}[ A{D^2} + {\left[ {\frac{{BC}}{2}} \right]^2} + BC.DM + \\A{D^2}+ {\left[ {\frac{{BC}}{2}} \right]^2} - BC.DM \end{Bmatrix} A{C^2} + A{B^2} \\&= 2A{D^2} + 2{\left[ {\frac{{BC}}{2}} \right]^2}A{C^2} + A{B^2} \\&= 2A{D^2} + \frac{1}{2}{B{C^2}} \end{align}\]

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