Question

D and E are points are on the sides CA and CB respectively of a triangle ABC right angles at C. Prove that

(AE) ^2 + (BD) ^2 = (AB) ^2 + (DE) ^2 °

Answer 1

In ∆ACE, by Pythagoras theorem,

$= > AE^2 = AC^2 + CE^2 \: \: \: \: \: \: \: \: ...( \textit{i})$


In ∆BCD, by Pythagoras theorem,

$= > BD^2 = CB^2 + DC^2 \: \: \: \: \: ...( \textit{ii})$


In ∆DCE, by Pythagoras theorem,

$= > DE^2 = DC^2 + CE^2 \: \: \: \: \: ...( \textit{iii})$


In ∆ABC, by Pythagoras theorem,

$= > AB^2 = AC^2 + CB^2 \: \: \: \: \: \:... ( \textit{iv})$



$\textit{Now, subtracting the sum of ( iii ) and ( iv ) } \\ \textit{from the sum of ( i ) and ( ii ), }$

= > AE² + BD² - ( DE² + AB² ) = AC^2 + CE^2 + CB² + DC² - ( DC² + CE² + AC² + CB² )


= > AE² + BD² - ( DE² + AB² ) = AC² + CE² + CB² + DC² - DC² - CE² - AC² - CB²


= > AE² + BD² - ( DE² + AB² ) = 0


= > AE² + BD² = DE² + AB²




Hence, proved.

Answer 2

Step-by-step explanation:

By Pythagoras theorem

(i) In ΔACE,

⇒ AC² + CE² = AE².

(ii) In ΔBCD,

⇒ BC² + CD² = DB²

From (i) & (ii), we have

AC² + CE² + BC² + CD² = AE² + DB²

(iii) In ΔCDE,

⇒ DE² = CD² + CE²

(iv) In ΔABC,

⇒ AB² = AC² + CB²

From (iii),(iv),(v), we have

DE² + AB² = AE² + DB²

Hope it helps you!