MCQ
If $ABC$ is an isosceles triangle and $D$ is a point on $BC$ such that $\text{AD}\perp\text{BC},$ then:
- A$AB^2 - AD^2 = BD \times DC$
- B$AB^2 - AD^2 = BD^2 - DC^2$
- ✓$AB^2 + AD^2 = BD \times DC$
- D$AB^2 + AD^2 = BD^2 - DC^2$
✓
Answer
Correct option: C.
$AB^2 + AD^2 = BD \times DC$
If $\triangle\text{ABC},$ $AB = AC$
$D$ is a point on $BC$ such that
$\text{AD}\perp\text{BC}$
$D$ is a point on $BC$ such that
$\text{AD}\perp\text{BC}$
$AD$ bisects $BC$ at $D$
In right $\triangle\text{ABD},$
$AB^2 = AD^2 + BD^2$
$AB^2 - AD^2 = BD^2 = BD \times BD = BD \times DC (BD = DC).$
In right $\triangle\text{ABD},$
$AB^2 = AD^2 + BD^2$
$AB^2 - AD^2 = BD^2 = BD \times BD = BD \times DC (BD = DC).$
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