Question : The sides AB, BC, and AC of a $\triangle {ABC}$ are 12 cm, 8 cm, and 10 cm respectively. A circle is inscribed in the triangle touching AB, BC, and AC at D, E, and F respectively. The difference between the lengths of AD and CE is:
Option 1: 4 cm
Option 2: 5 cm
Option 3: 3 cm
Option 4: 2 cm
Correct Answer: 4 cm
Solution :
Given: The sides AB, BC, and AC of a $\triangle {ABC}$ are 12 cm, 8 cm, and 10 cm respectively.
Tangents from a fixed point outside the circle always have the same length.
⇒ AD = AF, FC = CE, and BE = BD
Let the AD and CE be $x$ and $z$ respectively.
⇒ BE = BD = 12 - $x$
Also, CE = BC – BE
⇒ $z=8-12+x$
⇒ $x-z=4$
⇒ AD – CE = 4 cm
So, the difference between the lengths of AD and CE is 4 cm.
Hence, the correct answer is 4 cm.
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