In figure $ AD$ is a median of a triangle $ ABC$ and $ AM\perp BC$.
Prove that

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Solution

Step 1: Proving first condition
Given that,
In $∆ A B C$ , $A D$ is a median and $A M ⊥ B C$
In obtuse $∆ A D C$ , $∠ A D C > 90 °$ , therefore,
$A C^{2} = A D^{2} + D C^{2} + 2 \cdot D C \cdot D M \Rightarrow A C^{2} = A D^{2} + {(\frac{B C}{2})}^{2} + 2 \cdot (\frac{B C}{2}) \cdot D M [A D is median] \Rightarrow A C^{2} = A D^{2} + {(\frac{B C}{2})}^{2} + B C \cdot D M . . . . . . . . . . . . . . . . . . . . . (1)$
Step 2: Proving second condition
In acute $∆ A B D$ , $∠ A D M < 90 °$ , therefore,
$A B^{2} = A D^{2} + B D^{2} + 2 \cdot B D \cdot D M \Rightarrow A B^{2} = A D^{2} + {(\frac{B C}{2})}^{2} - 2 \cdot (\frac{B C}{2}) \cdot D M \Rightarrow A B^{2} = A D^{2} + {(\frac{B C}{2})}^{2} - B C \cdot D M . . . . . . . . . . . . (2)$
Step 3: Proving third condition
Now adding $(1)$ and $(2)$ we get
$A C^{2} + A B^{2} = 2 A D^{2} + 2 {(\frac{B C}{2})}^{2} \Rightarrow A C^{2} + A B^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}$
Hence, it is proved that
$A C^{2} = A D^{2} + B C \cdot D M + {(\frac{B C}{2})}^{2}$
$A B^{2} = A D^{2} - B C \cdot D M + {(\frac{B C}{2})}^{2}$
$A C^{2} + A B^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}$

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Similar questions

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i)

(ii)

(iii)

Seg AD is the median of △ ABC and AM ⊥ BC . Prove that (i) {AC}^{2} = {AD}^{2} + BC \times DM + {(\frac{BC}{2})}^{2} (ii) {AB}^{2} = {AD}^{2} + BC \times DM + {(\frac{BC}{2})}^{2}

{AB}^{2} = {AD}^{2} - BC \times DM + {(\frac{BC}{2})}^{2}

In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

(a) $B C^{2} + A D^{2} + 2 B C . A D$

(b) $A B^{2} + C D^{2} + 2 A B . C D$

(d) $B C^{2} + A D^{2} + 2 A B . C D$

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