MCQ
In a trapezium $ABCD$, if $AB \| CD$, then $(AC ^2 + BD ^2 ) = ?$
- A$BC ^2 + AD ^2 + 2BC ⋅ AD$
- B$AB^2 + CD^2+ 2AB ⋅ CD$
- C$AB^2 + CD^2 + 2AD ⋅ BC$
- ✓$BC^2 + AD^2 + 2AB ⋅ CD$
✓
Answer
Correct option: D.
$BC^2 + AD^2 + 2AB ⋅ CD$
Construction: Draw perpendicular from $D$ and $C$ on $A B$ which meets $A B$ at $E$ and $F$, respectively. So, $DEFC$ is a parallelogram, since one pair of opposite sides are parallel and equal.
In $\triangle\text{ABC},$
$\angle\text{B}$ is an acute angle.
$\Rightarrow\text{AC}^2=\text{BC}^2+\text{AB}^2-2\text{AB}\times\text{AE}$
In $\triangle\text{ABD},$
$\angle\text{A}$ is an acute angle.
$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$
$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$
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