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Polygons

In the given ...

Question

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

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Solution

Construction:
Join OA, OB and OC.
Applying Pythagoras theorem in $Δ A O F$ , we have
$O A^{2} = O F^{2} + A F^{2} . . . (i)$
Similarly, in $Δ B O D$
$O B^{2} = O D^{2} + B D^{2} . . . (i i)$
Similarly, in $Δ C O E$
$O C^{2} = O E^{2} + E C^{2} . . . (i i i)$
Adding the above equations, we get,
$O A^{2} + O B^{2} + O C^{2} = O F^{2} + A F^{2} + O D^{2} + B D^{2} + O E^{2} + E C^{2}$
$O A^{2} + O B^{2} + O C^{2} - O D^{2} - O E^{2} - O F^{2} = A F^{2} + B D^{2} + C E^{2}$ .

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Similar questions

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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}

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}

⊥

⊥

⊥

(i) 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}

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 B C

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⊥

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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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