Question
In a Trapezium $ABCD,$ if $\text{AB || CD},$ then $(AC^2 + BD^2) =$ ?
✓
Answer
Given: $ABCD$ is a trapezium with $\text{AB || CD}$
Construction: Draw $DE$ and $CF$ $\bot$ to $AB.$
Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore$ $BC^2 = AC^2 + AB^2 - 2AF : AB ...(1)$
and in $\triangle\text{BDA}$
$\angle\text{DBA}$ is acute
$\therefore$ $AD^2 = BD^2 + AB^2 - 2BE : AB ...(2)$
Adding $(1)$ and $(2)$ we get
$BC^2 + AD^2 = AC^2 + BD^2 + 2AB^2 - 2AF·AB - 2BE·AB$
$\Rightarrow AC^2 + BD^2 = BC^2 + AD^2 - 2AB [AB - AF - BE)$
$= BC^2 + AD^2 - 2AB [AB - (AE + EF) - (BF+ EF)]$
$= BC^2 + AD^2 - 2AB [AB - (AE + EF +BF+ EF)]$
$= BC^2 + AD^2 - 2AB [AB - (AB+ CD)] ( \therefore EF = DC)$
$= BC^2 + AD^2 - 2AB [- (CD)]$
$= AD^2 + BC^2 + 2AB \times CD$
Construction: Draw $DE$ and $CF$ $\bot$ to $AB.$
Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore$ $BC^2 = AC^2 + AB^2 - 2AF : AB ...(1)$
and in $\triangle\text{BDA}$
$\angle\text{DBA}$ is acute
$\therefore$ $AD^2 = BD^2 + AB^2 - 2BE : AB ...(2)$
Adding $(1)$ and $(2)$ we get
$BC^2 + AD^2 = AC^2 + BD^2 + 2AB^2 - 2AF·AB - 2BE·AB$
$\Rightarrow AC^2 + BD^2 = BC^2 + AD^2 - 2AB [AB - AF - BE)$
$= BC^2 + AD^2 - 2AB [AB - (AE + EF) - (BF+ EF)]$
$= BC^2 + AD^2 - 2AB [AB - (AE + EF +BF+ EF)]$
$= BC^2 + AD^2 - 2AB [AB - (AB+ CD)] ( \therefore EF = DC)$
$= BC^2 + AD^2 - 2AB [- (CD)]$
$= AD^2 + BC^2 + 2AB \times CD$
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