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Triangles

From a point ...

Question

From a point O in the interior of a $Δ A B C$ . Perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which one of the following is true?

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

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

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

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

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Solution

The correct option is A $A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}$

Assume triangle ABC to be an equilateral triangle. Let D, E, F be the mid-points for the sides BC, AC and AB respectively.
Then AF = FB = BD = DC = CE = EA.
Squaring and adding by making group of three will keep the equation equal.

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Similar questions

Δ 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}

△ 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}

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}

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}

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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}

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