Ex.6.5 Q8 Triangles Solution - NCERT Maths Class 10

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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 ]