In an equilateral \[\Delta \text{ }ABC,\text{ }D\]is a point on side BC such that \[BD=\frac{1}{3}BC\]. Prove that \[9{{(AD)}^{2}}\] \[=7{{(AB)}^{2}}\]. |
OR |
Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. |
Answer:
Given, ABC is an equilateral triangle and D is a point on BC such that \[BD=\frac{1}{3}BC\]. To prove: \[9A{{D}^{2}}=7A{{B}^{2}}\] Construction: Draw \[AE\bot BC\] Proof: \[BD=\frac{1}{3}BC\] ?(i) (Given) \[AE\bot BC\] We know that perpendicular from a vertex of equilateral triangle to the base divides base in two equal parts. \[\therefore \] \[BE=EC=\frac{1}{2}BC\] ?(ii) In \[\Delta \,AEB\], \[A{{D}^{2}}=A{{E}^{2}}+D{{E}^{2}}\] (Pythagoras theorem) Or \[A{{E}^{2}}=A{{D}^{2}}-D{{E}^{2}}\] ?(iii) Similarly, In \[\Delta \,AEB\], \[A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}\] \[=A{{D}^{2}}-D{{E}^{2}}+{{\left( \frac{1}{2}BC \right)}^{2}}\] [From (ii) and (iii)] \[=A{{D}^{2}}-{{(BE-BD)}^{2}}+\frac{1}{4}B{{C}^{2}}\] \[=A{{D}^{2}}-B{{E}^{2}}-B{{D}^{2}}+2.BE.BD+\frac{1}{4}B{{C}^{2}}\] \[=A{{D}^{2}}-{{\left( \frac{1}{2}BC \right)}^{2}}-{{\left( \frac{1}{3}BC \right)}^{2}}\]\[+2.\frac{1}{2}BC.\frac{1}{3}BC+\frac{1}{4}B{{C}^{2}}\] \[A{{B}^{2}}=A{{D}^{2}}-\frac{1}{9}B{{C}^{2}}+\frac{1}{3}B{{C}^{2}}\] \[\Rightarrow \] \[A{{B}^{2}}=A{{D}^{2}}+\frac{2}{9}B{{C}^{2}}\] \[\Rightarrow \] \[A{{B}^{2}}=A{{D}^{2}}+\frac{2}{9}A{{B}^{2}}\] (\[\because \] \[BC=AB\]) \[\Rightarrow \] \[A{{B}^{2}}-\frac{2}{9}A{{B}^{2}}=A{{D}^{2}}\] \[\Rightarrow \] \[\frac{7}{9}A{{B}^{2}}=A{{D}^{2}}\] \[\Rightarrow \] \[7A{{B}^{2}}=9A{{D}^{2}}\] Or \[9{{(AD)}^{2}}=7{{(AB)}^{2}}\] Hence Proved. OR Given: \[\Delta \text{ }ABC\]is a right angle triangle, right angled at A. To prove: \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] Construction: Draw \[AD\bot BC\]. Proof: In \[\Delta \text{ }ADB\]and \[\Delta \,BAC\], \[\angle B=\angle B\] (Common) \[\angle ADB=\angle BAC\] (Each \[90{}^\circ \]) \[\therefore \] \[\Delta \text{ }ADB\tilde{\ }\Delta \text{ }BAC\] (By AA similarity axiom) \[\therefore \] \[\frac{AB}{BC}=\frac{BD}{AB}\] (CPCT) \[A{{B}^{2}}=BC\times BD\] ?(i) Similarly, \[\Delta \text{ }ADC\tilde{\ }\Delta \text{ }CAB\] \[\frac{AC}{BC}=\frac{DC}{AC}\] \[A{{C}^{2}}=BC\times DC\] ?(ii) On adding eq. (i) and (ii) \[A{{B}^{2}}+A{{C}^{2}}=BC\times BD+BC\times CD\] \[=BC\text{ (}BD+CD)\] \[=BC\times BC\] \[A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\] \[\Rightarrow \] \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] Hence Proved.
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