# Ex.6.5 Q8 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.5'

## Question

In Figure, $$O$$ is a point in the interior of  $$\triangle ABC,$$ $$OD \bot BC, OE \bot AC\; \text{and}\;OF \bot AB$$. Show that

i. $$O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}-O{{D}^{2}}-O{{E}^{2}}-O{{F}^{2}}=AF{}^{2}+B{{D}^{2}}+C{{E}^{2}}$$

ii.  $$\,AF{}^{2}+B{{D}^{2}}+C{{E}^{2}}=AE{}^{2}+C{{D}^{2}}+B{{F}^{2}}$$

Diagram

## Text Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

(i) In $$\,\Delta ABC$$

$$OD\bot BC\,,\,OE\bot AC$$ and $$OF\bot AB$$

$OA , OB \text{and} \,OC \, \text{joined}$

In $$\,\Delta OAF$$

$$O{{A}^{2}}=AF{}^{2}+O{{F}^{2}}\;[\because \angle OFA={{90}^{0}}]$$………….(1)

Similarly, In $$\,\Delta OBD$$

$$O{{B}^{2}}=BD{}^{2}+O{{D}^{2}}\;[\because \angle ODA={{90}^{0}}]$$…………(2)

In $$\,\Delta OCE$$

$$\,O{{C}^{2}}=C{{E}^{2}}+O{{E}^{2}}\,\,\;\left[ \because \,\angle OEC={{90}^{\circ }} \right]$$………………(3)

Adding $$(1), (2)$$ and $$(3)$$

\begin{align} & O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=A{{F}^{2}}+O{{F}^{2}}+B{{D}^{2}}+O{{D}^{2}}+C{{E}^{2}}+O{{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}}.............(4) \\ \end{align}

(ii) From $$(4)$$

$(O{{A}^{2}}-O{{E}^{2}})+(O{{B}^{2}}-O{{F}^{2}})+(O{{C}^{2}}-O{{D}^{2}})=AF{}^{2}+B{{D}^{2}}+C{{E}^{2}}$

(Rearranging the left side terms)

$AE{}^{2}+B{{F}^{2}}+C{{D}^{2}}=AF{}^{2}+B{{D}^{2}}+C{{E}^{2}}$

[$$\because\Delta OAE,\Delta OBD$$ and $$\Delta OCE$$ are right triangles ]

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school