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Opposite & Adjacent Sides in a Right Angled Triangle

In the follow...

Question

In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA² + OB² + OC² − OD² − OE² − OF² = AF² + BD² + CE²

(ii) AF² + BD² + CE²= AE² + CD² + BF²

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Solution

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ΔAOF, we obtain

Similarly, in ΔBOD,

Similarly, in ΔCOE,

(ii) From the above result,

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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}

⊥

⊥

⊥

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 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}

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, 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}

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