A) Isosceles
B) Equilateral
C) Scalene
D) None of these
Correct Answer: A
Solution :
In \[\Delta ADB\]and\[\Delta ADC\] \[\angle BAD=\angle CAD\] (\[\because \]AD bisects \[\angle A\]) \[\angle ADB=\angle ADC\] (\[\because \] Each\[{{90}^{o}}\]) AD = AD (common) \[\Delta ADB\cong \Delta ADC\] (ASA congruency) \[\therefore \]AB = AC (by (PCT)) \[\therefore \]\[\Delta ABC\]is an isosceles triangle.You need to login to perform this action.
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