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Similar Triangles

In the given ...

Question

In the given fig O is a point in the interior of a triangle of a triangle $A B C, O D ⊥ B C, O E ⊥ A C$ and $O F ⊥ A B$ 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

REF.IMAGE.

Given: $Δ A B C$ and O is a point in thw inteeior of a $ \Delta ABC,
$w h e r e O D ⊥ B C, O E ⊥ A C, O F ⊥ A B .$

proof:-

Join the point O from A,B and C.

Using pythagoras theorem

In rt.angled $Δ l e O A F, O A^{2} = O F^{2} + A F^{2} . . . (1)$
In rt. angled $Δ l e O D B, . O B^{2} = O D^{2} + B D^{2} . . . (2)$
In rt. angled $Δ l e O E C, . O C^{2} = O E^{2} + E C^{2} . . . (3)$

Adding (1),(2) & (3)

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

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

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}

⊥

⊥

⊥

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

⊥

⊥

⊥

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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