Question

Given ABCD is a rhombus; $DE=BF$

Prove: AECF is a rhombus.

Short answer

As ABCD is a rhombus and diagonals of a rhombus are perpendicular bisectors of each other. So, $AO=CO$and $BO=DO$. Also, $AC\perp BD$. As $BO=DO$, subtract DE from both sides and use given $DE=BF$as shown below

$\begin{array}{c}BO-DE=DO-DE\\ BO-BF=DO-DE\\ FO=EO\end{array}$

Since, $AO=CO$and $FO=EO$. Also, AC are EF perpendicular to each other. So diagonals are perpendicular bisectors of each other. Thus, quadrilateral AECF is a rhombus.

Step-by-step solution

Step 1. Observation from given

As ABCD is a rhombus and diagonals of a rhombus are perpendicular bisectors of each other. So, $AO=CO$and $BO=DO$. Also, $AC\perp BD$.

Step 2. Show that EO=FO

As $BO=DO$, subtract DE from both sides and use given $DE=BF$as shown below

$\begin{array}{c}BO-DE=DO-DE\\ BO-BF=DO-DE\\ FO=EO\end{array}$

Step 3. Show AECF is a rhombus

Since, $AO=CO$and $FO=EO$. Also,$AC\text{and}EF$are perpendicular to each other. So diagonals are perpendicular bisectors of each other. Thus, quadrilateral AECF is a rhombus.