Math / Geometry

Question5. In the figure, $A B C D$ is a parallelogram. $A C$ cuts $B D$ at $E$ . $B C$ is produced to $F$ such that $F D=2 B E . \angle A D B=59^{\circ}$ and $\angle B A D=106^{\circ}$ . (a) Prove that $B D=F D$ . (b) Find $x$ and $y$ .
Solution: (a)

Studdy Solution

STEP 1

1. $ABCD$ is a parallelogram, which implies opposite sides are equal and opposite angles are equal.
2. $AC$ and $BD$ are diagonals of the parallelogram and intersect at point $E$ .
3. $BC$ is extended to $F$ such that $FD = 2BE$ .
4. $\angle ADB = 59^\circ$ and $\angle BAD = 106^\circ$ .

STEP 2

1. Use properties of parallelograms and triangles to prove $BD = FD$ .
2. Use angle properties to find $x$ and $y$ .

STEP 3

Since $ABCD$ is a parallelogram, opposite sides are equal. Thus, $AB = CD$ and $AD = BC$ .

STEP 4

The diagonals of a parallelogram bisect each other. Therefore, $BE = ED$ and $AE = EC$ .

STEP 5

Given that $FD = 2BE$ , and since $BE = ED$ , we have $FD = 2ED$ .

STEP 6

Since $FD = 2ED$ and $BE = ED$ , we conclude that $FD = BD$ .
(a) We have proved that $BD = FD$ .
(b) To find $x$ and $y$ , we need to use the given angles and properties of the parallelogram.

STEP 7

In $\triangle ABD$ , use the angle sum property. We know: $\angle ADB + \angle BAD + \angle ABD = 180^\circ$ Substitute the known angles: $59^\circ + 106^\circ + \angle ABD = 180^\circ$

STEP 8

Solve for $\angle ABD$ : $\angle ABD = 180^\circ - 59^\circ - 106^\circ = 15^\circ$

STEP 9

Since $ABCD$ is a parallelogram, opposite angles are equal. Therefore, $\angle BCD = \angle ABD = 15^\circ$ .

STEP 10

To find $x$ and $y$ , use the fact that $\angle ADB = 59^\circ$ and $\angle BCD = 15^\circ$ . Since $BC$ is extended to $F$ , $\angle BCF = \angle BCD = 15^\circ$ .
The values of $x$ and $y$ depend on additional information or context not provided in the problem statement. If $x$ and $y$ refer to specific angles or segments, please provide further details for accurate calculation.

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Question5. In the figure, $A B C D$ is a parallelogram. $A C$ cuts $B D$ at $E$ . $B C$ is produced to $F$ such that $F D=2 B E . \angle A D B=59^{\circ}$ and $\angle B A D=106^{\circ}$ . (a) Prove that $B D=F D$ . (b) Find $x$ and $y$ .
Solution: (a)

Studdy Solution

STEP 1

1. $ABCD$ is a parallelogram, which implies opposite sides are equal and opposite angles are equal.
2. $AC$ and $BD$ are diagonals of the parallelogram and intersect at point $E$ .
3. $BC$ is extended to $F$ such that $FD = 2BE$ .
4. $\angle ADB = 59^\circ$ and $\angle BAD = 106^\circ$ .

STEP 2

1. Use properties of parallelograms and triangles to prove $BD = FD$ .
2. Use angle properties to find $x$ and $y$ .

STEP 3

Since $ABCD$ is a parallelogram, opposite sides are equal. Thus, $AB = CD$ and $AD = BC$ .

STEP 4

The diagonals of a parallelogram bisect each other. Therefore, $BE = ED$ and $AE = EC$ .

STEP 5

Given that $FD = 2BE$ , and since $BE = ED$ , we have $FD = 2ED$ .

STEP 6

Since $FD = 2ED$ and $BE = ED$ , we conclude that $FD = BD$ .
(a) We have proved that $BD = FD$ .
(b) To find $x$ and $y$ , we need to use the given angles and properties of the parallelogram.

STEP 7

In $\triangle ABD$ , use the angle sum property. We know: $\angle ADB + \angle BAD + \angle ABD = 180^\circ$ Substitute the known angles: $59^\circ + 106^\circ + \angle ABD = 180^\circ$

STEP 8

Solve for $\angle ABD$ : $\angle ABD = 180^\circ - 59^\circ - 106^\circ = 15^\circ$

STEP 9

Since $ABCD$ is a parallelogram, opposite angles are equal. Therefore, $\angle BCD = \angle ABD = 15^\circ$ .

STEP 10

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Studdy Solution

STEP 1

STEP 2

1. Use properties of parallelograms and triangles to prove BD=FDBD = FDBD=FD.2. Use angle properties to find xxx and yyy.

STEP 3

Since ABCDABCDABCD is a parallelogram, opposite sides are equal. Thus, AB=CDAB = CDAB=CD and AD=BCAD = BCAD=BC.

STEP 4

The diagonals of a parallelogram bisect each other. Therefore, BE=EDBE = EDBE=ED and AE=ECAE = ECAE=EC.

STEP 5

Given that FD=2BEFD = 2BEFD=2BE, and since BE=EDBE = EDBE=ED, we have FD=2EDFD = 2EDFD=2ED.

STEP 6

Since FD=2EDFD = 2EDFD=2ED and BE=EDBE = EDBE=ED, we conclude that FD=BDFD = BDFD=BD. (a) We have proved that BD=FDBD = FDBD=FD.(b) To find xxx and yyy, we need to use the given angles and properties of the parallelogram.

STEP 7

In △ABD\triangle ABD△ABD, use the angle sum property. We know: ∠ADB+∠BAD+∠ABD=180∘\angle ADB + \angle BAD + \angle ABD = 180^\circ∠ADB+∠BAD+∠ABD=180∘ Substitute the known angles: 59∘+106∘+∠ABD=180∘59^\circ + 106^\circ + \angle ABD = 180^\circ59∘+106∘+∠ABD=180∘

STEP 8

Solve for ∠ABD\angle ABD∠ABD: ∠ABD=180∘−59∘−106∘=15∘\angle ABD = 180^\circ - 59^\circ - 106^\circ = 15^\circ∠ABD=180∘−59∘−106∘=15∘

STEP 9

Since ABCDABCDABCD is a parallelogram, opposite angles are equal. Therefore, ∠BCD=∠ABD=15∘\angle BCD = \angle ABD = 15^\circ∠BCD=∠ABD=15∘.

STEP 10

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1. Use properties of parallelograms and triangles to prove $BD = FD$ .
2. Use angle properties to find $x$ and $y$ .

Since $ABCD$ is a parallelogram, opposite sides are equal. Thus, $AB = CD$ and $AD = BC$ .

The diagonals of a parallelogram bisect each other. Therefore, $BE = ED$ and $AE = EC$ .

Given that $FD = 2BE$ , and since $BE = ED$ , we have $FD = 2ED$ .

Since $FD = 2ED$ and $BE = ED$ , we conclude that $FD = BD$ .
(a) We have proved that $BD = FD$ .
(b) To find $x$ and $y$ , we need to use the given angles and properties of the parallelogram.

In $\triangle ABD$ , use the angle sum property. We know: $\angle ADB + \angle BAD + \angle ABD = 180^\circ$ Substitute the known angles: $59^\circ + 106^\circ + \angle ABD = 180^\circ$

Solve for $\angle ABD$ : $\angle ABD = 180^\circ - 59^\circ - 106^\circ = 15^\circ$

Since $ABCD$ is a parallelogram, opposite angles are equal. Therefore, $\angle BCD = \angle ABD = 15^\circ$ .