Question

In a right angled $△ABC$, right angled at $A$, if $AD\perp BC$ such that $AD=p$, if $BC=a,CA=b$ and $AB=c$, then:

• A. $p^2=b^2+c^2$
• B. $\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2}$
• C. $\frac{p}{a}=\frac{p}{b}$
• D. $p^2=b^2{c}^2$

Answer 1

The correct option is B $\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2}$

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, ABC and DAC

$\angle ABC=\angle CAD$ ..(From III)

$\angle BAC=\angle CDA$ (Each ${90}^{\circ })$

$\angle ACD=\angle ACB$ (Common angle)

Thus, $△ABC\sim △DAC$ (AAA rule)

Thus, $\frac{AC}{DC}=\frac{BC}{AC}$ (Sides of similar triangles are in proportion)

$A{C}^2=BC\times DC$... (V)
Similarly, $△ABC\sim △DBA$

and $\frac{AB}{BD}=\frac{BC}{AB}$ (Sides of similar triangles are in proportion)

$A{B}^2=BD\times BC$ ....(VI)
Similarly, $△ABD\sim △CAD$ (AAA rule)

Thus, $\frac{AD}{CD}=\frac{BD}{AD}$ (Sides of similar triangles are in proportion)

$A{D}^2=BD\times CD$
Now,

$\frac{1}{A{B}^2}+\frac{1}{A{C}^2}=\frac{1}{BC\times DC}+\frac{1}{BD\times BC}$

$\frac{1}{A{B}^2}+\frac{1}{A{C}^2}=\frac{1}{BC}(\frac{1}{DC}+\frac{1}{BD})$

$\frac{1}{A{B}^2}+\frac{1}{A{C}^2}=\frac{1}{BC}\left(\frac{BD+DC}{BD\times DC}\right)$

$\frac{1}{A{B}^2}+\frac{1}{A{C}^2}=\frac{1}{BC}\left(\frac{BC}{BD\times DC}\right)$

$\frac{1}{A{B}^2}+\frac{1}{A{C}^2}=\frac{1}{BD\times DC}$

$\frac{1}{A{B}^2}+\frac{1}{A{C}^2}=\frac{1}{A{D}^2}$

$\frac{1}{c^2}+\frac{1}{b^2}=\frac{1}{p^2}$