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Visualising Pythagoras Theorem

Question 8 ii...

Question

Question 8 (ii)
In the given figure, O is a point in the interior of a triangle ABC, OD $⊥$ BC, OE $⊥$ AC and OF $⊥$ AB.
Show that (ii) $A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}$

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Solution

(ii) Join OA, OB and OC

We have, $O A^{2} + O B^{2} + O C^{2} - O D^{2} - O E^{2} - O F^{2} = A F^{2} + B D^{2} + C E^{2}$
$A F^{2} + B D^{2} + E C^{2} = (O A^{2} - O E^{2}) + (O C^{2} - O D^{2}) + (O B^{2} - O F^{2})$
$∴ A F^{2} + B D^{2} + C E^{2} = A E^{2} + C D^{2} + B F^{2}$

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Similar questions

Q. Question 8 (ii)
In the given figure, O is a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB.
Show that (ii) \( AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2\)

Q. In Figure, O is a point in the interior of a triangle ABC, OD

⊥

⊥

⊥

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

.

O

is a point in the interior of a triangle

A B C, O D ⊥ B C, O E ⊥ A C

O F ⊥ A B

. Show that
(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}

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

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

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

Q. In given figure from a point O in the interior of a

△ A B C

, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :

{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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Visualising Pythagoras Theorem

Standard IX Mathematics

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