In Figure, O is a point in the interior of a triangle ABC, OD $⊥$ BC, OE $⊥$ AC and OF $⊥$ AB. Show that $A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}$ .

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Solution

From the all the right angled triangles, we get
$O A^{2} = A F^{2} + O F^{2}$
$O A^{2} = A E^{2} + O E^{2}$
$O B^{2} = O D^{2} + B D^{2}$
$O B^{2} = O F^{2} + F B^{2}$
$O C^{2} = O E^{2} + E C^{2}$
$O C^{2} = O D^{2} + C D^{2}$
We have,
$A F^{2} + B D^{2} + C E^{2} = O A^{2} - O F^{2} + O B^{2} - O D^{2} + O C^{2} - O E^{2}$
$⟹ (O A^{2} - O E^{2}) + (O B^{2} - O F^{2}) + (O C^{2} - O D^{2})$
$⟹ A E^{2} + C D^{2} + B F^{2}$ . Hence proved.

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Similar questions

O

A B C, O D ⊥ B C, O E ⊥ A C

O F ⊥ A B

O A^{2} + O B^{2} + O C^{2} - O D^{2} - O E^{2} - O F^{2} = A F^{2} + B D^{2} + C E^{2}

A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}

A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}

⊥

⊥

⊥

A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}

△ A B C

{A F}^{2} + {B D}^{2} + {C E}^{2} = {O A}^{2} + {O B}^{2} + {O C}^{2} - {O D}^{2} - {O E}^{2} - {O F}^{2}

{A F}^{2} + {B D}^{2} + {C E}^{2} = {A E}^{2} + {C D}^{2} + {B F}^{2}

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