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In the given figure, AB || DE and BD || EF. Prove that $DC^2$ = CF x АС
In the given figure, AB || DE and BD || EF. Prove that $DC^2$ = CF x АС
This Question has 1 answers.
Since $AB \parallel DE$ and $BD \parallel EF$, triangles formed by these parallel lines with transversals are similar by the basic proportionality theorem (Thales' theorem).
Using similarity properties in $\triangle DCF$ and $\triangle ACD$:
Since $\triangle DCF \sim \triangle ACD$, we have the property of similar triangles:
$ \frac{DC}{AC} = \frac{CF}{DC} $
Cross multiplying:
$ DC^2 = CF \times AC $
$\boxed{DC^2 = CF \times AC}$
Using similarity properties in $\triangle DCF$ and $\triangle ACD$:
Since $\triangle DCF \sim \triangle ACD$, we have the property of similar triangles:
$ \frac{DC}{AC} = \frac{CF}{DC} $
Cross multiplying:
$ DC^2 = CF \times AC $
$\boxed{DC^2 = CF \times AC}$
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