Difference between revisions of "Ceva's theorem/Problems"

Latest revision as of 16:16, 9 May 2021

Introductory

I1

Problem

Suppose $AB, AC$ , and $BC$ have lengths $13, 14$ , and $15$ , respectively. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$ , find $BD$ and $DC$ .

Solution

If $BD = x$ and $DC = y$ , then $10x = 40y$ , and ${x + y = 15}$ . From this, we find $x = 12$ and $y = 3$ .

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Difference between revisions of "Ceva's theorem/Problems"

Latest revision as of 16:16, 9 May 2021

Contents

Introductory

I1

Problem

Solution

Latest revision as of 16:16, 9 May 2021 (view source) Etmetalakret (talk \| contribs) ← Older edit	Latest revision as of 16:16, 9 May 2021 (view source) Etmetalakret (talk \| contribs)
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