Question

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Figure-1

ABC is a right angle triangle and ∠BAC=90°

So, $BC^{2}=AB^{2}+AC^{2}$...................equation 1

In ΔABC; AD, BE and CF are medians.

The length of the medians is AD

$AD=\frac{1}{2} \sqrt{2AC^{2}+2AB^{2}-BC^{2}}\\\\AD^{2}=\frac{1}{4}(2AC^{2}+2AB^{2}-BC^{2})\\\\4AD^{2}=2AC^{2}+2AB^{2}-BC^{2}\\\\4AD^{2}=2BC^{2}-BC^{2}\\\\4AD^{2}=BC^{2}$ ..............equation 2

In ΔBAE, ∠BAE=90°

We know that In a triangle, a median is a line joining a vertex with the mid-point of the opposite side.

$\therefore{AE}=\frac{1}{2}AC$

$BE^{2}=AB^{2}+AE^{2}\\\\BE^{2}=AB^{2}+(\frac{1}{2}AC)^{2}\\\\BE^{2}=AB^{2}+\frac{1}{4}AC^{2}\\\\4BE^{2}=4AB{2}+AC^{2}$...... equation 3

In ΔCAF, ∠CAF=90°

We know that In a triangle, a median is a line joining a vertex with the mid-point of the opposite side.

$\therefore{AF}=\frac{1}{2}AB$

$CF^{2}=AC^{2}+AF^{2}\\\\CF^{2}=AC^{2}+(\frac{1}{2}AB)^{2}\\\\CF^{2}=AC^{2}+\frac{1}{4}AB^{2}\\\\4CF^{2}=4AC{2}+AB^{2}$...... equation 4

Now adding the equation 2, 3 and 4 we get

$4AD^{2}+4BE^{2}+4CF^{2}=BC^{2}+4AB^{2}+AC^{2}+4AC^{2}+AB^{2}\\\\4AD^{2}+4BE^{2}+4CF^{2}=5AB^{2}+5AC^{2}+BC^{2}\\\\4AD^{2}+4BE^{2}+4CF^{2}=5BC^{2}+BC^{2} (\text{Put the value of equation 1})\\\\4AD^{2}+4BE^{2}+4CF^{2}=6BC^{2}\\\\2AD^{2}+4BE^{2}+4CF^{2}=3BC^{2}$

Hence the RHS = LHS (Proved)