OACB is a quadrilateral. OB is parallel to AC. D is the point on BC such that $BD = \frac{1}{3} BC$ . The lines OB and AD produced meet at F. Explain why $FB = \dfrac{1}{2} AC$ .

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Worked solution

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1. Find the ratio $BD:DC$ from the given fraction of $BC$.

Given
$BD=\frac13BC.$
Hence
$DC=BC-BD=BC-\frac13BC=\frac23BC.$
So
$BD:DC=\frac13BC:\frac23BC=1:2.$

2. Establish that the two relevant triangles are similar.

Since $FB$ lies on the same straight line as $OB$ , and $OB \parallel AC$ ,
$FB \parallel AC.$
Also, $B,D,C$ are collinear and $F,D,A$ are collinear.

Therefore
$\angle FBD=\angle ACD$
and
$\angle FDB=\angle ADC.$
So
$\triangle FBD \sim \triangle ACD.$

3. Use corresponding sides of similar triangles to prove the required length relationship.

For similar triangles, corresponding sides are proportional:
$\frac{FB}{AC}=\frac{BD}{DC}.$
Since $BD:DC=1:2$ ,
$\frac{FB}{AC}=\frac12.$
Therefore
$FB=\frac12AC.$

Answer

Since $BD=\dfrac13BC$ , $DC=\dfrac23BC$ , so $BD:DC=1:2$ . Also, $FB \parallel AC$ , with $B,D,C$ collinear and $F,D,A$ collinear. Hence $\triangle FBD$ is similar to $\triangle ACD$ . Therefore
$\frac{FB}{AC}=\frac{BD}{DC}=\frac{1}{2},$
so
$FB=\frac12 AC.$

AI analysis — may contain mistakes.

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OACB is a quadrilateral. OB is parallel to AC. D is the point on BC such that BD=13BCBD = \frac{1}{3} BCBD=31​BC. The lines OB and AD produced meet at F. Explain why FB=12ACFB = \dfrac{1}{2} ACFB=21​AC.

Worked solution

7

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OACB is a quadrilateral. OB is parallel to AC. D is the point on BC such that $BD = \frac{1}{3} BC$ . The lines OB and AD produced meet at F. Explain why $FB = \dfrac{1}{2} AC$ .