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Pythagoras Theorem

ii In the giv...

Question

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) \( AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2\)

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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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Q. Question 8 (ii)
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Pythagoras Theorem

Standard X Mathematics

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