## Online Course Discussion Forum

### Question about 6.28 II-A Geometry

Hello, I am reviewing my missed questions and on this question, I am not sure about the part where the answer provided make sense.

It is the part where it says triangle EBD is similar to triangle BEF.

Can I get some help, please.

thank you

Re: Question about 6.28 II-A Geometry

Also I need some help on 6.29 as well, I don't get what is going on

Re: Question about 6.28 II-A Geometry

Here we are using the property that if two triangles are similar, and the ratio of their sides is $a : b$, then the ratio of their areas is $a^2 : b^2$.

So, for example, since $\triangle AFG \sim \triangle ADE$ and the ratio of their sides is $AF : AD$, then the ratio of their areas is $AF^2 : AD^2$.

Re: Question about 6.28 II-A Geometry

Yeah I get that part, but how to prove that they are similar is  my question

Re: Question about 6.28 II-A Geometry

This is because $DE$ and $FG$ are parallel to $BC$. The parallelnes of the segments gives you angles of the same size, so by AA the triangles are similar.

Re: Question about 6.28 II-A Geometry

$AC$ is parallel to $BE$, and $AC \perp BD$, so $\angle EBD = 90^\circ$.

Both $\triangle EBD$ and $\triangle BFD$ are right and share $\angle D$, so they must be similar.