Question

ABC is an isosceles triangle with AB $=$AC and D is a point on BC such that $AD\perp BC$ (Fig. 7.13). To prove that $\angle BAD=\angle CAD,$ a student proceeded as follows:
$\Delta ABD$ and $\Delta ACD,$
AB $=$ AC (Given)
$\angle B=\angle C$ (because AB $=$ AC)
and $\angle ADB=\angle ADC$
Therefore, $\Delta ABD\cong \Delta ACD\left(AAS\right)$
So, $\angle BAD=\angle CAD\left(CPCT\right)$
What is the defect in the above arguments?

• A. It is defective to use $\angle ABD=\angle ACD$ for proving this result.
• B. It is defective to use $\angle ADB=\angle ADC$ for proving this result.
• C. It is defective to use $\angle BAD=\angle DCA$ for proving this result.
• D. Cannot be determined
  

Answer 1

Answer: (A)

$△ABD$ and $△ACD$

AB=AC (given )

Then $\angle ABD=\angle ACD$ ( because AB=AC )

and $\angle ADB=\angle ADC=90$( because AD⊥BC )

$\therefore △ABD=△ACD$

$\angle BAD=\angle CAD$

It is defective to use $\angle ABD=\angle ACD$ for proving this result