question_answer

\[\Delta \text{ }ABC\] is a right angled triangle in which \[\angle B=90{}^\circ \]. D and E are any point on AB and BC respectively. Prove that\[A{{E}^{2}}+C{{D}^{2}}=A{{C}^{2}}+D{{E}^{2}}\].

Answer:

In \[\Delta \text{ }ABC,\angle B=60{}^\circ \] and D, E are point of AB, BC respectively.

To prove:

\[A{{C}^{2}}+D{{E}^{2}}=A{{E}^{2}}+C{{D}^{2}}\]

In \[\Delta \text{ }ABC\] by using Pythagoras theorem,

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]                                     ?(i)

In \[\Delta \text{ }ABE\] by using Pythagoras theorem

\[A{{E}^{2}}=A{{B}^{2}}+B{{E}^{2}}\]                           ?(ii)

In \[\Delta \text{ }BCD\]by Pythagoras theorem

\[C{{D}^{2}}=B{{D}^{2}}+B{{C}^{2}}\]                         ?(iii)

In \[\Delta \text{ }DBF\] by Pythagoras theorem

\[D{{E}^{2}}=D{{B}^{2}}+B{{E}^{2}}\]                         ?(iv)

Adding eq. (i) and eq. (iv)

\[A{{C}^{2}}+D{{E}^{2}}=A{{B}^{2}}+B{{C}^{2}}+B{{D}^{2}}+B{{E}^{2}}\]

\[=A{{B}^{2}}+B{{E}^{2}}+B{{C}^{2}}+B{{D}^{2}}\]

\[A{{C}^{2}}+D{{E}^{2}}=A{{E}^{2}}+C{{D}^{2}}\]

[From eq. (ii) and eq. (iii)]                       Hence Proved.