Question在 $\square A B C D$ 中, $A C, B D$ 交于点 $O$ , 点 $E, F$ 在 $A C$ 上, $A E=C F$ 。证明四边形 $E B F D$ 是平行四边形和菱形。

Studdy Solution

STEP 1

Assumptions1. We have a square $ABCD$ . . Diagonals $AC$ and $BD$ intersect at point $.<br />3. Points$ and $$ are on diagonal $AC$ such that $AE=CF$.
4. We need to prove that quadrilateral $EBFD$ is a parallelogram.
5. If $\angle BAC = \angle DAC$, we need to prove that quadrilateral $EBFD$ is a rhombus.

STEP 2

In order to prove that quadrilateral $EBFD$ is a parallelogram, we need to show that opposite sides are equal or parallel.

STEP 3

Since $AE=CF$ and $AO=OC$ (as $$ is the intersection point of diagonals of square $ABCD$), we can say that triangles $AEO$ and $CFO$ are congruent by the Side-Side-Side (SS) criterion.
$\triangle AEO \cong \triangle CFO$

STEP 4

From congruent triangles, we know that corresponding parts of congruent triangles are equal. Therefore, we have $\angle AEO = \angle CFO$ and \angleAO = \angle FCO.

STEP 5

Since $\angle AEO = \angle CFO$ and \angleAO = \angle FCO, we can say that \angleAB = \angle FCB and $\angle BAE = \angle BCF$ .

STEP 6

As \angleAB = \angle FCB and $\angle BAE = \angle BCF$ , we can say that \angleBD = \angle FBD and $\angle BED = \angle BFD$ .

STEP 7

Since \angleBD = \angle FBD and $\angle BED = \angle BFD$ , we can say that triangles $EB$ and $BD$ are congruent by the Angle-Side-Angle (ASA) criterion.
\triangleBD \cong \triangle FBD

STEP 8

From congruent triangles, we know that corresponding parts of congruent triangles are equal. Therefore, we have $EB = FB$ and $BD = BD$ .

STEP 9

Since $EB = FB$ and $BD = BD$ , we can say that $EBFD$ is a parallelogram because both pairs of opposite sides are equal.

STEP 10

Now, if $\angle BAC = \angle DAC$ , we can say that $\angle BAE = \angle DCF$ .

STEP 11

Since $\angle BAE = \angle DCF$ , we can say that $\angle BED = \angle BFD$ .

STEP 12

As $\angle BED = \angle BFD$ , we can say that $ED = FD$ .

STEP 13

Since $ED = FD$ and we have already proven that $EB = FB$ , we can say that $EBFD$ is a rhombus because all sides are equal.
The quadrilateral $EBFD$ is a parallelogram and if $\angle BAC = \angle DAC$ , then $EBFD$ is a rhombus.

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