Question

State true or false:

In $\Delta ABC,\angle A={90}^{\circ }$ and $AD\perp BC.$ Then, $A{B}^2+A{C}^2=B{C}^2.$

• A. True
• B. False

Answer 1

The correct option is A True

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)
Adding (V) and (VI)
$A{C}^2+A{B}^2=BC\times DC+BD\times BC$
$A{C}^2+B{C}^2=BC(DC+BD)$
$A{C}^2+B{C}^2=BC\times BC$
$A{C}^2+B{C}^2=B{C}^2$