Elegance is the key - Part II

Geometry Level 3

In the above figure, points $(E,D,F )$ and $(B,A,F)$ are collinear, and $\overline{EC} = \overline{CD} = \overline{BC}$ . Also, $CB \bot BA$ . The length of $ED$ is $a$ , and the length of $DF$ is $b$ . Then find the length of $FB$ in terms of $a$ and $b$ .

This is an original problem.

\sqrt{(a+b)a}

(a^{2} - b^{2})

\sqrt{(a-b)b}

\sqrt{(a+b)b}

\sqrt{ab}

1 solution

Venkata Karthik Bandaru
Jul 10, 2015

Since $\overline{EC} = \overline{CD} = \overline{CB}$ , i.e. the points $D, B$ and $E$ are equidistant from point $C$ , it implies that a circle can be drawn passing through through $E, D$ and $B$ with centre $C$ .

Now, as $EDF$ and $BAF$ are straight lines and $CB \bot BF$ , we clearly see that $BF$ is a tangent, and $EF$ is a secant to the circle $C$ . Hence, from Tangent-Secant Theorem , we have $\overline{FB}^{2} = \overline{FE} \cdot \overline{FD} \Rightarrow \boxed{\overline{FB} = \sqrt{(a+b)b}}$

Moderator note:

Great approach! Nice way to hide the fact that we have a circle :)

Damn! I wrongly used Tangent secant theorem and ended up clicking $\sqrt{ab}$ . I had almost cracked the whole problem!

Nihar Mahajan - 5 years, 11 months ago

You could have checked your answer with a special case when $a=0.$ The length of $FB$ would be equal to $b$ . Certainly $\sqrt{ab}$ doesn't qualify, in fact there's only one option satisfying the criteria, leaving all the others out.

Sanjeet Raria - 5 years, 11 months ago

hi karthik I am shivanand I think I am not good at geometry so please suggest me books and websites or advise.

shivanand biradar - 5 years, 11 months ago

Elegance is the key - Part II

This is an original problem.

1 solution

Moderator note:

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