Math / Geometry

Question(k) Prove that bisectors of any two adjacent angles of a parallelogram are at right (ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel. (iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square. (i) If ABCD is a rectangle in which the diagonal BD bisects $\angle \mathrm{B}$ , then show that $A B C D$ is a square. (ii) Show that if the diagonals of a quadrilateral are equal and bisect each other a

Studdy Solution

STEP 1

What is this asking? We're diving into the world of parallelograms, exploring angle bisectors and proving some cool properties about rectangles and squares! Watch out! Don't mix up adjacent and opposite angles – they're key to unlocking this geometric puzzle!
Also, remember all the properties that make a square a square.

STEP 2

1. Adjacent Angle Bisectors
2. Opposite Angle Bisectors
3. Equal Diagonals, Right Angle Bisectors
4. Rectangle with Diagonal Bisecting an Angle
5. Equal Diagonals and Bisection

STEP 3

Let's imagine a parallelogram $ABCD$ .
We'll draw bisectors of adjacent angles $\angle A$ and $\angle B$ .
Let these bisectors intersect at point $E$ .

STEP 4

In parallelogram $ABCD$ , we know that adjacent angles are supplementary.
So, $\angle A + \angle B = 180^\circ$ .

STEP 5

Since $AE$ and $BE$ are angle bisectors, we have $\angle EAB = \frac{1}{2} \angle A$ and $\angle EBA = \frac{1}{2} \angle B$ .

STEP 6

Now, look at the triangle $ABE$ .
The sum of its angles is $180^\circ$ .
Thus, $\angle EAB + \angle EBA + \angle AEB = 180^\circ$ .

STEP 7

Substitute the values from 2.1.3 into the equation from 2.1.4: $\frac{1}{2} \angle A + \frac{1}{2} \angle B + \angle AEB = 180^\circ$ .

STEP 8

Multiply both sides by $2$ : $\angle A + \angle B + 2 \cdot \angle AEB = 360^\circ$ .

STEP 9

From 2.1.2, we know $\angle A + \angle B = 180^\circ$ . Substitute this into the equation from 2.1.6: $180^\circ + 2 \cdot \angle AEB = 360^\circ$ .

STEP 10

Subtract $180^\circ$ from both sides: $2 \cdot \angle AEB = 180^\circ$ .

STEP 11

Divide both sides by $2$ : $\angle AEB = 90^\circ$ .
Boom! The bisectors are perpendicular!

STEP 12

In parallelogram $ABCD$ , opposite angles are equal: $\angle A = \angle C$ and $\angle B = \angle D$ .

STEP 13

Let $AE$ and $CF$ be bisectors of $\angle A$ and $\angle C$ , respectively.
Then, $\angle EAB = \frac{1}{2} \angle A$ and $\angle FCD = \frac{1}{2} \angle C$ .

STEP 14

Since $\angle A = \angle C$ , we have $\frac{1}{2} \angle A = \frac{1}{2} \angle C$ , which means $\angle EAB = \angle FCD$ .

STEP 15

These equal angles are corresponding angles formed by the transversal $AC$ intersecting lines $AE$ and $CF$ .
Therefore, $AE \parallel CF$ .
The bisectors are parallel!

STEP 16

If diagonals are equal and bisect at right angles, we have a rhombus.
Since diagonals bisect each other, all sides are equal.
Since they bisect at right angles, we have four right angles.
Equal sides and all right angles?
That's a square!

STEP 17

In rectangle $ABCD$ , diagonal $BD$ bisects $\angle B$ .
So, $\angle ABD = \angle CBD$ .

STEP 18

Since $ABCD$ is a rectangle, $\angle A = \angle C = 90^\circ$ .

STEP 19

In $\triangle ABD$ and $\triangle CBD$ , $AD = BC$ (opposite sides of a rectangle), $BD = BD$ (common side), and $\angle ABD = \angle CBD$ (given).
By SAS congruence, $\triangle ABD \cong \triangle CBD$ .

STEP 20

Therefore, $AB = BC$ .
A rectangle with adjacent sides equal is a square!

STEP 21

If diagonals are equal and bisect each other, the quadrilateral is a rectangle.
If they also bisect each other at right angles, it's a rhombus.
A rectangle that's also a rhombus is a square!

STEP 22

We've proven that adjacent angle bisectors in a parallelogram are perpendicular, opposite angle bisectors are parallel, a quadrilateral with equal diagonals bisecting at right angles is a square, and a rectangle with a diagonal bisecting an angle is a square.
We've also shown that a quadrilateral with equal diagonals that bisect each other is a rectangle, and if they bisect at right angles, it's a square!

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Adjacent Angle Bisectors2. Opposite Angle Bisectors3. Equal Diagonals, Right Angle Bisectors4. Rectangle with Diagonal Bisecting an Angle5. Equal Diagonals and Bisection

STEP 3

Let's **imagine** a parallelogram ABCDABCDABCD.We'll **draw** bisectors of adjacent angles ∠A\angle A∠A and ∠B\angle B∠B.Let these bisectors **intersect** at point EEE.

STEP 4

In parallelogram ABCDABCDABCD, we know that adjacent angles are supplementary.So, ∠A+∠B=180∘\angle A + \angle B = 180^\circ∠A+∠B=180∘.

STEP 5

Since AEAEAE and BEBEBE are angle bisectors, we have ∠EAB=12∠A\angle EAB = \frac{1}{2} \angle A∠EAB=21​∠A and ∠EBA=12∠B\angle EBA = \frac{1}{2} \angle B∠EBA=21​∠B.

STEP 6

Now, look at the triangle ABEABEABE.The sum of its angles is 180∘180^\circ180∘.Thus, ∠EAB+∠EBA+∠AEB=180∘\angle EAB + \angle EBA + \angle AEB = 180^\circ∠EAB+∠EBA+∠AEB=180∘.

STEP 7

**Substitute** the values from 2.1.3 into the equation from 2.1.4: 12∠A+12∠B+∠AEB=180∘\frac{1}{2} \angle A + \frac{1}{2} \angle B + \angle AEB = 180^\circ21​∠A+21​∠B+∠AEB=180∘.

STEP 8

**Multiply** both sides by 222: ∠A+∠B+2⋅∠AEB=360∘\angle A + \angle B + 2 \cdot \angle AEB = 360^\circ∠A+∠B+2⋅∠AEB=360∘.

STEP 9

From 2.1.2, we know ∠A+∠B=180∘\angle A + \angle B = 180^\circ∠A+∠B=180∘. **Substitute** this into the equation from 2.1.6: 180∘+2⋅∠AEB=360∘180^\circ + 2 \cdot \angle AEB = 360^\circ180∘+2⋅∠AEB=360∘.

STEP 10

**Subtract** 180∘180^\circ180∘ from both sides: 2⋅∠AEB=180∘2 \cdot \angle AEB = 180^\circ2⋅∠AEB=180∘.

STEP 11

**Divide** both sides by 222: ∠AEB=90∘\angle AEB = 90^\circ∠AEB=90∘.Boom! The bisectors are perpendicular!

STEP 12

In parallelogram ABCDABCDABCD, opposite angles are equal: ∠A=∠C\angle A = \angle C∠A=∠C and ∠B=∠D\angle B = \angle D∠B=∠D.

STEP 13

Let AEAEAE and CFCFCF be bisectors of ∠A\angle A∠A and ∠C\angle C∠C, respectively.Then, ∠EAB=12∠A\angle EAB = \frac{1}{2} \angle A∠EAB=21​∠A and ∠FCD=12∠C\angle FCD = \frac{1}{2} \angle C∠FCD=21​∠C.

STEP 14

Since ∠A=∠C\angle A = \angle C∠A=∠C, we have 12∠A=12∠C\frac{1}{2} \angle A = \frac{1}{2} \angle C21​∠A=21​∠C, which means ∠EAB=∠FCD\angle EAB = \angle FCD∠EAB=∠FCD.

STEP 15

These equal angles are **corresponding angles** formed by the transversal ACACAC intersecting lines AEAEAE and CFCFCF.Therefore, AE∥CFAE \parallel CFAE∥CF.The bisectors are parallel!

STEP 16

If diagonals are equal and bisect at right angles, we have a rhombus.Since diagonals bisect each other, all sides are equal.Since they bisect at right angles, we have four right angles.Equal sides and all right angles?That's a square!

STEP 17

In rectangle ABCDABCDABCD, diagonal BDBDBD bisects ∠B\angle B∠B.So, ∠ABD=∠CBD\angle ABD = \angle CBD∠ABD=∠CBD.

STEP 18

Since ABCDABCDABCD is a rectangle, ∠A=∠C=90∘\angle A = \angle C = 90^\circ∠A=∠C=90∘.

STEP 19

In △ABD\triangle ABD△ABD and △CBD\triangle CBD△CBD, AD=BCAD = BCAD=BC (opposite sides of a rectangle), BD=BDBD = BDBD=BD (common side), and ∠ABD=∠CBD\angle ABD = \angle CBD∠ABD=∠CBD (given).By SAS congruence, △ABD≅△CBD\triangle ABD \cong \triangle CBD△ABD≅△CBD.

STEP 20

Therefore, AB=BCAB = BCAB=BC.A rectangle with adjacent sides equal is a square!

STEP 21

If diagonals are equal and bisect each other, the quadrilateral is a rectangle.If they also bisect each other at right angles, it's a rhombus.A rectangle that's also a rhombus is a square!

STEP 22

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Adjacent Angle Bisectors
2. Opposite Angle Bisectors
3. Equal Diagonals, Right Angle Bisectors
4. Rectangle with Diagonal Bisecting an Angle
5. Equal Diagonals and Bisection

Let's imagine a parallelogram $ABCD$ .
We'll draw bisectors of adjacent angles $\angle A$ and $\angle B$ .
Let these bisectors intersect at point $E$ .

In parallelogram $ABCD$ , we know that adjacent angles are supplementary.
So, $\angle A + \angle B = 180^\circ$ .

Since $AE$ and $BE$ are angle bisectors, we have $\angle EAB = \frac{1}{2} \angle A$ and $\angle EBA = \frac{1}{2} \angle B$ .

Now, look at the triangle $ABE$ .
The sum of its angles is $180^\circ$ .
Thus, $\angle EAB + \angle EBA + \angle AEB = 180^\circ$ .

Substitute the values from 2.1.3 into the equation from 2.1.4: $\frac{1}{2} \angle A + \frac{1}{2} \angle B + \angle AEB = 180^\circ$ .

Multiply both sides by $2$ : $\angle A + \angle B + 2 \cdot \angle AEB = 360^\circ$ .

From 2.1.2, we know $\angle A + \angle B = 180^\circ$ . Substitute this into the equation from 2.1.6: $180^\circ + 2 \cdot \angle AEB = 360^\circ$ .

Subtract $180^\circ$ from both sides: $2 \cdot \angle AEB = 180^\circ$ .

Divide both sides by $2$ : $\angle AEB = 90^\circ$ .
Boom! The bisectors are perpendicular!

In parallelogram $ABCD$ , opposite angles are equal: $\angle A = \angle C$ and $\angle B = \angle D$ .

Let $AE$ and $CF$ be bisectors of $\angle A$ and $\angle C$ , respectively.
Then, $\angle EAB = \frac{1}{2} \angle A$ and $\angle FCD = \frac{1}{2} \angle C$ .

Since $\angle A = \angle C$ , we have $\frac{1}{2} \angle A = \frac{1}{2} \angle C$ , which means $\angle EAB = \angle FCD$ .

These equal angles are corresponding angles formed by the transversal $AC$ intersecting lines $AE$ and $CF$ .
Therefore, $AE \parallel CF$ .
The bisectors are parallel!

If diagonals are equal and bisect at right angles, we have a rhombus.
Since diagonals bisect each other, all sides are equal.
Since they bisect at right angles, we have four right angles.
Equal sides and all right angles?
That's a square!

In rectangle $ABCD$ , diagonal $BD$ bisects $\angle B$ .
So, $\angle ABD = \angle CBD$ .

Since $ABCD$ is a rectangle, $\angle A = \angle C = 90^\circ$ .

In $\triangle ABD$ and $\triangle CBD$ , $AD = BC$ (opposite sides of a rectangle), $BD = BD$ (common side), and $\angle ABD = \angle CBD$ (given).
By SAS congruence, $\triangle ABD \cong \triangle CBD$ .

Therefore, $AB = BC$ .
A rectangle with adjacent sides equal is a square!

If diagonals are equal and bisect each other, the quadrilateral is a rectangle.
If they also bisect each other at right angles, it's a rhombus.
A rectangle that's also a rhombus is a square!