Question

$A{D}^2=BD\times DC$

If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given, In $△ABC$, $\angle A=90$ and $AD\perp BC$
In $△ABC$,
$\angle BAC+\angle ABC+\angle ACB=180$
$90+\angle ABC+\angle ACB=180$
$\angle ABC+\angle ACB=90$ (I)
In $△CAD$,
$\angle CAD+\angle ACD+\angle ADC=180$
$\angle CAD+\angle ACD+90=180$
$\angle CAD+\angle ACD=90$..(II)
Equating (I) and (II),
$\angle ABC+\angle ACB=\angle CAD+\angle ACD$
$\angle ABC=\angle CAD$...(III)
Similarly, $\angle ACB=\angle BAD$...(IV)
Now, In $△$s, ABD and CAD
$\angle ABD=\angle CAD$...(From III)
$\angle BAD=\angle ACB$ ..(From IV)
$\angle ADB=\angle ADC$ (Each ${90}^{\circ })$
Thus, $△ABD\sim △CAD$ (AAA rule)
Thus, $\frac{AD}{CD}=\frac{BD}{AD}$ (Sides of similar triangles are in proportion)
$A{D}^2=BD\times CD$