# 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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