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Right Triangles Having Common Edges

In the given ...

Question

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

True

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False

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Solution

The correct option is B False
Join OA, OB and OC.

Applying Pythagoras theorem in $Δ A O F$ , we have,
$O A^{2} = O F^{2} + A F^{2}$
Similarly, in $Δ B O D$ , $O B^{2} = O D^{2} + B D^{2}$
Similarly, in $Δ C O E$ , $O C^{2} = O E^{2} + E C^{2}$
Adding these equations,
$O A^{2} + O B^{2} + O C^{2} = O F^{2} + A F^{2} + O D^{2} + B D^{2} + O E^{2} + E C^{2}$
$O A^{2} + O B^{2} + O C^{2} - O D^{2} - O E^{2} - O F^{2} = A F^{2} + B D^{2} + C E^{2}$

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⊥

⊥

⊥

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^{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

{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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