Question

given isosceles triangle abc with ab ≅ bc. point d is the midpoint of ab. e is the midpoint of bc, and f is the midpoint of ca. use a similarity transformation and triangle congruence to show that △abc~△fec.
dilate △abc with c as center and scale factor 0.5. by the definition of dilation, ac = ( )ac and bc = ( )bc. dilation preserves angle measure, so m∠acb = by the definition of midpoint, and by the transitive property of equality, and so ac≅fc and bc≅ec. so, △abc≅△fec by therefore, △abc~△fec.

Explanation:

Step1: Recall dilation property

When dilating $\triangle ABC$ with center $C$ and scale - factor $0.5$, by the definition of dilation, $A'C' = 0.5AC$ and $B'C'=0.5BC$.

Step2: Note angle - preservation of dilation

Dilation preserves angle measure, so $m\angle A'C'B'=m\angle ACB$.

Step3: Use mid - point definition

Since $F$ is the mid - point of $CA$, $FC = 0.5AC$ and since $E$ is the mid - point of $BC$, $EC = 0.5BC$.

Step4: Apply transitive property

By the transitive property of equality, $A'C'=FC$ and $B'C' = EC$.

Step5: Determine triangle congruence

So, $\triangle A'B'C'\cong\triangle FEC$ by the Side - Angle - Side (SAS) congruence criterion. Since $\triangle A'B'C'$ is a dilation of $\triangle ABC$, $\triangle ABC\sim\triangle FEC$.

Answer:

$A'C' = 0.5AC$, $B'C' = 0.5BC$, $m\angle A'C'B'=m\angle ACB$, $FC = 0.5AC$, $EC = 0.5BC$, $A'C' = FC$, $B'C' = EC$, SAS