MCQ
In $\triangle\text{ABC}\angle\text{BAC}=90^\circ$ and $\text{AD} \bot\text{BC}$ then :
- A$\text{BD}.\text{CD}=\text{BC}^2$
- B$\text{AB}.\text{AC}=\text{BC}^2$
- ✓$\text{BD}.\text{CD}=\text{AD}^2$
- D$\text{AB}.\text{AC}=\text{AD}^2$
✓
Answer
Correct option: C.
$\text{BD}.\text{CD}=\text{AD}^2$
In $\triangle\text{ABC}$ and $\triangle\text{ADC}$
$\angle\text{D}=\angle\text{D}=90^\circ$
$\angle\text{DBA}=\angle\text{DAC}$
By $\text{AAA}$ similarity criterion,
$\triangle\text{ADB}\sim\triangle\text{ADC}$
$\frac{\text{BD}}{\text{AD}}=\frac{\text{AD}}{\text{CD}}$
$\text{BD.CD}=\text{AD}^2$
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