Question

euclidean geometry
step by step process

1. (7 pts) Prove that the diagonals of a rectangle are congruent. 2. (18 pts) In the diagram below, prove that M is the midpoint of AC and BD if and only if ABCD is a parallelogram. 3. (9 pts) Use #1 and #2 above to prove that the diagonals of a square cut each other into 4 congruent segments. Use #3 to prove that the diagonals of a square are angle bisectors of its right 4" (8 pts) angles. 5, (8 pts) Use #4 to prove the diagonals of a square are perpendicular to each other.

Answer 1

1.

Consider the rectangle ABCD with diagonals AC and BD

to prove AC and BD are congruent

Consider $\bigtriangleup$ ABC and $\bigtriangleup$ BDC

AB$\cong$ DC [opposite sides of a rectangle are congruent]

$\angle$ B $\cong \angle$ C [ each angle of a rectangle = 90⁰]

BC $\cong$ BC [common side]

$\bigtriangleup$ ABC $\cong \bigtriangleup$ BDC [by SAS axiom]

AC $\cong$ BD [corresponding parts of congruent triangles are congruent]

Therefore diagonals are congruent.

2

Consider the given figure ABCD in the question

Let M be the midpoint of AC and BD

TO prove ABCD is aparallelogram

Consider$\bigtriangleup$ DMC and $\bigtriangleup$AMB

DM $\cong$ MB [since M is the midpoint of BD

AM$\cong$ MC [since M is the midpoint of AC

$\angle$ DMC$\cong$$\angle$AMB [ vertical angles are equal]

$\bigtriangleup$ DMC $\cong$$\bigtriangleup$ AMB [by SAS axiom]

$\angle$ MDC$\cong$$\angle$ MBA [corresponding parts of congruent triangles are congruent]

then AB parallel CD because$\angle$ MDC and $\angle$MBA are alternate interior angles

Also AB = CD [corresponding parts of congruent triangles]

then ABCD is a parallelogram because a pair of opposite sides are parallel and equal

Now assume ABCD is a parallelogram

To prove M is the midpoint of AC and BD

Consider $\bigtriangleup$ CMD and $\bigtriangleup$ AMB

$\angle$ MDC$\cong$$\angle$ MBA[ alternate interior angles are equal as AB and CD are parallel]

$\angle$MCD$\cong$$\angle$ MAB [alternate interior angles]

AB$\cong$ CD [opposite sides of a parallelogram]

$\bigtriangleup$ CMD$\cong$$\bigtriangleup$ AMB[by SAS axiom]

AM =MC and BM = MD[ corresponding parts of congruent triangles a congruent]

then M is the midpoint of AC and BD

3

Square is a rectangle with all sides are equal

then by (1) diagonals of a square are congruent

Squareis a parallelogram with equal sides and equal angles

then by (2) diagonals of a square intersect at midpoint

so by (1) and (2) diagonals of a square cut each other into 4 congruent segments

4

Consider square ABCD and the diagonals AC and BD meet at O

by(3) AO = OC =BO=OD

Consider $\bigtriangleup$ AOB and $\bigtriangleup$ BOC

AO$\cong$ OC [given]

AB$\cong$ BC [sides of the square]

BO $\cong$ BO[common side]

$\bigtriangleup$ AOB $\cong$$\bigtriangleup$ BOC [by SSS axiom]

$\angle$OBA $\cong$$\angle$OBC [corresponding parts of congruent triangles are congruent]

that means BD bisects $\angle$ B

Similar is the casewith other angles .

Hence diagonals of a square are angle bisectors.

Answer 2

1.

Consider the rectangle ABCD with diagonals AC and BD

to prove AC and BD are congruent

Consider $\bigtriangleup$ ABC and $\bigtriangleup$ BDC

AB$\cong$ DC [opposite sides of a rectangle are congruent]

$\angle$ B $\cong \angle$ C [ each angle of a rectangle = 90⁰]

BC $\cong$ BC [common side]

$\bigtriangleup$ ABC $\cong \bigtriangleup$ BDC [by SAS axiom]

AC $\cong$ BD [corresponding parts of congruent triangles are congruent]

Therefore diagonals are congruent.

2

Consider the given figure ABCD in the question

Let M be the midpoint of AC and BD

TO prove ABCD is aparallelogram

Consider$\bigtriangleup$ DMC and $\bigtriangleup$AMB

DM $\cong$ MB [since M is the midpoint of BD

AM$\cong$ MC [since M is the midpoint of AC

$\angle$ DMC$\cong$$\angle$AMB [ vertical angles are equal]

$\bigtriangleup$ DMC $\cong$$\bigtriangleup$ AMB [by SAS axiom]

$\angle$ MDC$\cong$$\angle$ MBA [corresponding parts of congruent triangles are congruent]

then AB parallel CD because$\angle$ MDC and $\angle$MBA are alternate interior angles

Also AB = CD [corresponding parts of congruent triangles]

then ABCD is a parallelogram because a pair of opposite sides are parallel and equal

Now assume ABCD is a parallelogram

To prove M is the midpoint of AC and BD

Consider $\bigtriangleup$ CMD and $\bigtriangleup$ AMB

$\angle$ MDC$\cong$$\angle$ MBA[ alternate interior angles are equal as AB and CD are parallel]

$\angle$MCD$\cong$$\angle$ MAB [alternate interior angles]

AB$\cong$ CD [opposite sides of a parallelogram]

$\bigtriangleup$ CMD$\cong$$\bigtriangleup$ AMB[by SAS axiom]

AM =MC and BM = MD[ corresponding parts of congruent triangles a congruent]

then M is the midpoint of AC and BD

3

Square is a rectangle with all sides are equal

then by (1) diagonals of a square are congruent

Squareis a parallelogram with equal sides and equal angles

then by (2) diagonals of a square intersect at midpoint

so by (1) and (2) diagonals of a square cut each other into 4 congruent segments

4

Consider square ABCD and the diagonals AC and BD meet at O

by(3) AO = OC =BO=OD

Consider $\bigtriangleup$ AOB and $\bigtriangleup$ BOC

AO$\cong$ OC [given]

AB$\cong$ BC [sides of the square]

BO $\cong$ BO[common side]

$\bigtriangleup$ AOB $\cong$$\bigtriangleup$ BOC [by SSS axiom]

$\angle$OBA $\cong$$\angle$OBC [corresponding parts of congruent triangles are congruent]

that means BD bisects $\angle$ B

Similar is the casewith other angles .

Hence diagonals of a square are angle bisectors.