In Fig. $6.54, \mathrm{O}$ is a point in the interior of a triangle $\mathrm{ABC}, \mathrm{OD} \perp \mathrm{BC}, \mathrm{OE} \perp \mathrm{AC}$ and $\mathrm{OF} \perp \mathrm{AB}$. Show that
(i) $\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}$,
(ii) $\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}=\mathrm{AE}^{2}+\mathrm{CD}^{2}+\mathrm{BF}^{2}$.

Answer

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Geometry

NCERT Class 10 - Math

Chapter 6

Triangles

Section 5

Areas of Similar Triangles

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Video Transcript

Hello. We have problem number eight questions. Is that in figure all in the point in the interior angle, in terms of a triangle? Abc ODed perpendicular to this? This this And so we have to show thanks, we tell being given. So let us draw the figure and find out. Okay, Yeah, this is A B and C. And this is perpendicular to this. Okay, There's a perpendicular, this is perpendicular E. Ah And D. And O is being given. Uh First of all, we need to prove over a squared plus B squared plus b squared minus four squared minus the square mile square square. So this is being given to prove. And second, we can also prove in the similar way. Okay. First is oh, what we need to prove is or is quite place will be a squared plus B squared. So no problem. We have to do some construction. So let us construct this. There's and Uh huh. This. Okay, now we have 12 and three and various right angled triangles because all are perpendicular. So let us right try right angle triangle O. F. These are right angled triangles strangle B or D. And triangle C. O. E. Will be having oh a square equal to a f square place. We have a square. Similarly for triangle bot. Oh be square will be equal to bt square place. Oh the square and thurday's from triangle C O A O C square place. Well, this is quite equal to see a square place. E. It's quiet. So we will be just adding it up. It will be always Squire, bliss will be squared plus who sees quite equal to a F square place? Oh F square plus B. D squared plus O. D squared plus C. E. Square place, E squared. Now we needed to prove that there is quite a bit square. Cisco minus all the square. So these things. Right, okay, so from here, yeah. Okay, from here we should Right Oh, a square place or be square plus or c square minus. Let's take Odie. Oh, odie Oef and oif This side. Who F esquire. Okay, minus all the square minus. We square equal to a f square place. BbD square place. Ce square these things. This is the question. We need to prove I needed to be approved. Okay, second is I wanted to prove again. Well, certain things. So and we can say that will be a square minus O C square equal to b D square miners cd esquire. And similarly oc square minus who is square equal to ceo square minus it is square and last Who a square minus will be square equal to f square minus BF square. Now, if we added up these things will get cancelled out. So the left hand side will become zero zero equal to be the Squire minus cd Squire place. See a square minus a squared plus f square minus BF square. Okay, so if you take these things are left side, it will come be sbd square place. Ce esquire unless you have a square equal to uh this is a c D squared plus a square place. BF squad pretends fruit. Thank you.

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