MCQ
In an isosceles triangle $\text{ABC}$ if $AC = BC$ and $AB^2= 2AC^2,$ then $\angle\text{C}=$
- A$30^\circ$
- B$45^\circ$
- ✓$90^\circ$
- D$60^\circ$
✓
Answer
Correct option: C.
$90^\circ$
In isosceles $\triangle\text{ABC},\ \text{AC}=\text{BC}$
and $A B^2=A C^2+A C^2=2 A C^2$
$=A C^2+B C^2(A C=B C)$
By converse of Pythagoras
Theorem $, \angle\text{C}=90^\circ$
and $A B^2=A C^2+A C^2=2 A C^2$
$=A C^2+B C^2(A C=B C)$
By converse of Pythagoras
Theorem $, \angle\text{C}=90^\circ$
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