# In the figure, O is a point in the interior of a triangle ABC, OD $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ AB.

## Question :

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

(i)  $${OA}^2$$ + $${OB}^2$$ + $${OC}^2$$ – $${OD}^2$$ – $${OE}^2$$ – $${OF}^2$$ = $${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$

(ii)  $${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$ = $${AE}^2$$ + $${CD}^2$$ + $${BF}^2$$

## Solution :

Join AO, BO and CO as shown in fig.

(i)  In right $$\triangle$$s OFA, ODB and OEC, we have

$${OA}^2$$ = $${AF}^2$$ + $${OF}^2$$

$${OB}^2$$ = $${BD}^2$$ + $${OD}^2$$

and $${OC}^2$$ = $${CE}^2$$ + $${OE}^2$$

$${OA}^2$$ + $${OB}^2$$ + $${OC}^2$$ = $${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$ + $${OD}^2$$ + $${OE}^2$$ + $${OF}^2$$

or   $${OA}^2$$ + $${OB}^2$$ + $${OC}^2$$ – $${OD}^2$$ – $${OE}^2$$ – $${OF}^2$$ = $${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$

(ii)  In right $$\triangle$$s ODB and ODC, we have :

$${OB}^2$$ = $${BD}^2$$ + $${OD}^2$$  and  $${OC}^2$$ = $${OD}^2$$ + $${CD}^2$$

or  $${OB}^2$$ – $${OC}^2$$ = $${BD}^2$$ – $${CD}^2$$            ……….(1)

Similarly, we have :

$${OC}^2$$ – $${OA}^2$$ = $${CE}^2$$ – $${AE}^2$$               ………..(2)

$${OA}^2$$ – $${OB}^2$$ = $${AF}^2$$ + $${BF}^2$$              ………..(3)

Adding equations (1), (2) and (3), we get

$${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$ = $${AE}^2$$ + $${CD}^2$$ + $${BF}^2$$