The perimeter of $\triangle A B C$ is 32 . If $\angle A B C=\angle A C B$ and $B C=12$, the length of $A B$ is
(A) 20
(B) 12
(C) 10
(D) 8
(E) 16

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-060.jpg?height=309&width=456&top_left_y=274&top_left_x=1233)
Since $\angle A B C=\angle A C B$, then $\triangle A B C$ is isosceles and $A B=A C$.

Given that $\triangle A B C$ has a perimeter of $32, A B+A C+12=32$ or $A B+A C=20$.

But $A B=A C$, so $2 A B=20$ or $A B=10$.

ANSWER: $(\mathrm{C})$