triangle similarity problem.

Question

$2ac=bc$ find the ratio ( $K$ ) what is the ratio of their area? I found out it is $2$ or $1/2$ is it true?

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Answer 1

Let's call the area of left triangle $A_1$ and the right triangle $A_2$.

A quick guide:

As the angles: $\angle ADC = \angle BYC$, all angles of triangle are 2 by 2 equal so the triangles are similar (not equal) because $\dfrac{AC}{BC}=\dfrac{1}{2}$, so the ratio of each vertex are equal to $2$ or $1/2$ hence $\dfrac{A_1}{A_2}=\big(\dfrac{1}{2}\big)^2=\dfrac{1}{4}$

The comprehensive solution:

Since they both are right-angled triangles, we have:

$\displaystyle A_1=\frac{AC \times CD}{2}$

$\displaystyle A_2=\frac{BC \times CY}{2}$

$\displaystyle \dfrac{A_1}{A_2}=\dfrac{\frac{AC \times CD}{2}}{\frac{BC \times CY}{2}}=\frac{AC}{BC}\frac{CD}{CY}=\frac{CD}{2\times CY}$

$K=\dfrac{A_1}{A_2}$ or $\dfrac{A_2}{A_1}$

Now if $CD=CY$ then $K=\dfrac12$ or $2$, but $\require{cancel} CD\cancel{=}CY!$

As the angles: $\angle ADC = \angle BYC$, the triangles are similar (not equal), hence $\dfrac{AC}{BC}=\dfrac{CD}{CY}=\dfrac{1}{2}$

$K=\dfrac{CD}{2\times CY}=\dfrac{1}{4}$ or $\dfrac{2\times CY}{CD}=4$

Answer 2

You marked the angles at $f$ and $g$ (or $d$ and $y$) are equal. So the triangles are similar.
If one has vertical edge half the other, their bases will also be in ratio 2 to 1.