After a long break

Geometry Level 2

In $\triangle ABC$ , $AD$ is the median from $A$ ; point $E$ on $AD$ is such that $\dfrac {AE}{AD} = \dfrac 12$ ; and the extension of $BE$ meets $AC$ at $F$ . Find $\dfrac {AF}{FC}$ .

\frac{1}{6}

\frac{1}{8}

\frac{1}{4}

\frac{1}{2}

\frac{1}{5}

\frac{1}{3}

1 solution

Aryan Sanghi
Sep 10, 2020

Use Mass point Geometry .

Thanks upvoted. I have my phase test tomorrow so can you give a solution .I have no time,and no idea about mass point theorem

SRIJAN Singh - 9 months ago

I'll post today when I'll get time.

Aryan Sanghi - 9 months ago

@Aryan Sanghi , @Kriti Kamal Can you please tell how do I improve my geometry? I am very bad at it, I know very less theorems, and I can't solve any problems.

Vinayak Srivastava - 9 months ago

Don't worry I'm with you I recommend you this web that has all of the theorems check it out ,and for practice you have materials.Keep learning:)

SRIJAN Singh - 9 months ago

After a long break

1 solution

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