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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

In figure, A...

Question

In figure, $A D$ is a median of $△ A B C$ .Prove that $A B + A C > 2 A D$

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Solution

In $△ A D B$ and $△ E D C$
$A D = D E$ and $B D = D C$ since $A D$ is a median of $△ A B C$
$∠ A D B = ∠ E D C$ (vertically opposite angles)
$∴ △ A D B ≅ △ E D C$ by SAS axiom
$∴ A B = E C$ ....... $(1)$ by CPCT
In $△ A E C,$
$A C + E C > A E$
$∵$ The sum of any two sides of a triangle is greater than the third side.
$\Rightarrow A C + A B > A E$ from $(1)$
$\Rightarrow A B + A C > A E$
$\Rightarrow A B + A C > 2 A D$

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Similar questions

Δ A B C,

A B^{2} + A C^{2} = 2 A D^{2} + 2 B D^{2} ?

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A M ⊥ B C .

(i) A C^{2} = A D^{2} + B C \times D M + {(\frac{B C}{2})}^{2}

A D

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A M ⊥ B C .

(i i) A B^{2} = A D^{2} - B C \times D M + {(\frac{B C}{2})}^{2}

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A M ⊥ B C .

(i i i) A C^{2} + A B^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}

△ A B C, A D

A E ⊥ B C

A B^{2} + A C^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}

In $Δ A B C,$ if AD is a median, which of the following relations holds good?

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