Given: AB .BE = CB . BD Prove: Triangle ABC is similar to triangle DBE

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Given: AB .BE = CB . BD Prove: Triangle ABC is similar to triangle DBE

Geometry
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I don't understand the given. Could you please type it correctly?
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I have not had Geometry, I watched a video explaining AA SAS and SS
\[(AB)(BE)=(CB)(BD)\] \[\frac{(AB)(BE)}{(BE)(BD)}=\frac{(CB)(BD)}{(BE)(BD)}\] \[\frac{(AB)}{(BD)}=\frac{(CB)}{(BE)}\]
So now we have established that corresponding sides have equal ratios.
Now use the fact that vertical angles are congruent and you have SAS similarity.
I knew about vertical angles
thanks
yw

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