• # question_answer 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.

 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.