QuestionIn square $ABCD$ , points $E$ on $AB$ and $F$ on $BC$ satisfy $AF = DE$ . Show: (a) $\triangle ABF \cong \triangle DAE$ ; (b) Is $AF$ perpendicular to $DE$ ? Why? (c) Prove $\triangle ABF \sim \triangle AGE$ .

Studdy Solution

STEP 1

Assumptions1. $ABCD$ is a square. . $and$ are points on $AB$ and $BC$ respectively.
3. $AF = DE$ .
4. $AF$ and $DE$ intersect at $G$ .

STEP 2

To prove that $\triangle AB \cong \triangle DAE$ , we need to show that they have the same side lengths and angles.Since $ABCD$ is a square, we know that $AB = AD$ and $BC = CD$ .Given that $AF = DE$ , we can say that $AB$ and $DA$ have two sides of the same length.

STEP 3

Since $ABCD$ is a square, we know that $\angle BAF = \angle DAE =90^{\circ}$ .

STEP 4

Now, we can use the Side-Angle-Side (AS) congruence rule to prove that $\triangle AB \cong \triangle DAE$ .The SAS rule states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent.

STEP 5

For part (b), we need to determine if $AF$ is perpendicular to $DE$ .We know that $AF = DE$ and they intersect at $G$ . However, this information alone is not enough to determine if $AF$ is perpendicular to $DE$ .We would need additional information, such as the measurement of $\angle AG$ or $\angle FGE$ , to determine if $AF$ is perpendicular to $DE$ .

STEP 6

For part (c), to prove that $\triangle AB \sim \triangle AGE$ , we need to show that the ratios of the corresponding sides are equal and the corresponding angles are equal.

STEP 7

We know that $AB = AG$ (since $G$ is the intersection point of $AF$ and $DE$ ) and $AF = GE$ (given).So, the ratio of the corresponding sides $AB/AG = AF/GE =1$ .

STEP 8

Also, since $\triangle AB \cong \triangle DAE$ , we know that $\angle BAF = \angle DAE$ .
And since $ABCD$ is a square, we know that $\angle BAC = \angle DAG =90^{\circ}$ .
So, the corresponding angles are equal.

STEP 9

Therefore, by the Side-Side-Angle (SSA) similarity rule, we can say that $\triangle AB \sim \triangle AGE$ .
The SSA rule states that if two sides in a ratio and the included angle of one triangle are equal to two sides in a ratio and the included angle of another triangle, then the triangles are similar.

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QuestionIn square $ABCD$ , points $E$ on $AB$ and $F$ on $BC$ satisfy $AF = DE$ . Show: (a) $\triangle ABF \cong \triangle DAE$ ; (b) Is $AF$ perpendicular to $DE$ ? Why? (c) Prove $\triangle ABF \sim \triangle AGE$ .

Studdy Solution

STEP 1

Assumptions1. $ABCD$ is a square. . $and$ are points on $AB$ and $BC$ respectively.
3. $AF = DE$ .
4. $AF$ and $DE$ intersect at $G$ .

STEP 2

To prove that $\triangle AB \cong \triangle DAE$ , we need to show that they have the same side lengths and angles.Since $ABCD$ is a square, we know that $AB = AD$ and $BC = CD$ .Given that $AF = DE$ , we can say that $AB$ and $DA$ have two sides of the same length.

STEP 3

Since $ABCD$ is a square, we know that $\angle BAF = \angle DAE =90^{\circ}$ .

STEP 4

Now, we can use the Side-Angle-Side (AS) congruence rule to prove that $\triangle AB \cong \triangle DAE$ .The SAS rule states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent.

STEP 5

STEP 6

For part (c), to prove that $\triangle AB \sim \triangle AGE$ , we need to show that the ratios of the corresponding sides are equal and the corresponding angles are equal.

STEP 7

We know that $AB = AG$ (since $G$ is the intersection point of $AF$ and $DE$ ) and $AF = GE$ (given).So, the ratio of the corresponding sides $AB/AG = AF/GE =1$ .

STEP 8

Also, since $\triangle AB \cong \triangle DAE$ , we know that $\angle BAF = \angle DAE$ .
And since $ABCD$ is a square, we know that $\angle BAC = \angle DAG =90^{\circ}$ .
So, the corresponding angles are equal.

STEP 9

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QuestionIn square ABCDABCDABCD, points EEE on ABABAB and FFF on BCBCBC satisfy AF=DEAF = DEAF=DE. Show: (a) △ABF≅△DAE\triangle ABF \cong \triangle DAE△ABF≅△DAE; (b) Is AFAFAF perpendicular to DEDEDE? Why? (c) Prove △ABF∼△AGE\triangle ABF \sim \triangle AGE△ABF∼△AGE.

Studdy Solution

STEP 1

Assumptions1. ABCDABCDABCD is a square. . and and and are points on ABABAB and BCBCBC respectively.3. AF=DEAF = DEAF=DE.4. AFAFAF and DEDEDE intersect at GGG.

STEP 2

STEP 3

Since ABCDABCDABCD is a square, we know that ∠BAF=∠DAE=90∘\angle BAF = \angle DAE =90^{\circ}∠BAF=∠DAE=90∘.

STEP 4

STEP 5

STEP 6

For part (c), to prove that △AB∼△AGE\triangle AB \sim \triangle AGE△AB∼△AGE, we need to show that the ratios of the corresponding sides are equal and the corresponding angles are equal.

STEP 7

We know that AB=AGAB = AGAB=AG (since GGG is the intersection point of AFAFAF and DEDEDE) and AF=GEAF = GEAF=GE (given).So, the ratio of the corresponding sides AB/AG=AF/GE=1AB/AG = AF/GE =1AB/AG=AF/GE=1.

STEP 8

Also, since △AB≅△DAE\triangle AB \cong \triangle DAE△AB≅△DAE, we know that ∠BAF=∠DAE\angle BAF = \angle DAE∠BAF=∠DAE.And since ABCDABCDABCD is a square, we know that ∠BAC=∠DAG=90∘\angle BAC = \angle DAG =90^{\circ}∠BAC=∠DAG=90∘.So, the corresponding angles are equal.

STEP 9

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QuestionIn square $ABCD$ , points $E$ on $AB$ and $F$ on $BC$ satisfy $AF = DE$ . Show: (a) $\triangle ABF \cong \triangle DAE$ ; (b) Is $AF$ perpendicular to $DE$ ? Why? (c) Prove $\triangle ABF \sim \triangle AGE$ .

Assumptions1. $ABCD$ is a square. . $and$ are points on $AB$ and $BC$ respectively.
3. $AF = DE$ .
4. $AF$ and $DE$ intersect at $G$ .

Since $ABCD$ is a square, we know that $\angle BAF = \angle DAE =90^{\circ}$ .

For part (c), to prove that $\triangle AB \sim \triangle AGE$ , we need to show that the ratios of the corresponding sides are equal and the corresponding angles are equal.

We know that $AB = AG$ (since $G$ is the intersection point of $AF$ and $DE$ ) and $AF = GE$ (given).So, the ratio of the corresponding sides $AB/AG = AF/GE =1$ .

Also, since $\triangle AB \cong \triangle DAE$ , we know that $\angle BAF = \angle DAE$ .
And since $ABCD$ is a square, we know that $\angle BAC = \angle DAG =90^{\circ}$ .
So, the corresponding angles are equal.