Question

in the figure above, circle ( o ) is inscribed in ( \triangle abc ) so that the circle is tangent to ( overline{ab} ) at ( f ), to ( overline{bc} ) at ( e ), and to ( overline{ac} ) at ( d ). if ( ce = 12 ), ( eb = 7 ) and ( ab = 15 ), find ( ac ).
( \bigcirc ) ( 18 )
( \bigcirc ) ( 14 )
( \bigcirc ) ( 24 )
( \bigcirc ) ( 20 )

Explanation:

Step1: Recall tangent segment theorem

Tangents from a common external point to a circle are equal in length. So, \( CE = CD = 12 \), \( EB = FB = 7 \), and let \( AD = AF = x \).

Step2: Find length of \( AF \)

We know \( AB = AF + FB = 15 \) and \( FB = 7 \), so \( x + 7 = 15 \). Solving for \( x \), we get \( x = 15 - 7 = 8 \). Thus, \( AD = 8 \).

Step3: Calculate \( AC \)

Since \( AC = AD + CD \), and \( AD = 8 \), \( CD = 12 \), then \( AC = 8 + 12 = 20 \).

Answer:

20