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1 year ago

ABCD is a parallelogram. E is the point where the diagonals AC and BD meet. Prove that triangle ABE is congruent to triangle CDE

Mathematics

2 answers:

iogann1982 [59]1 year ago

7 0

Answer:

Step-by-step explanation:

Given that ABCD is a parallelogram.

E is the point where the diagonals AC and BD meet.

Consider triangles ABE and CDE

Statement Reason

--------------------------------------------------------------------

1) AB =CD Opposite sides of a parallelogram

2) Angle ABE =

Angle EDC Alternate angles of two parallel lines

3) BE = DE Diagonals bisect each other in a parallelogram

Hence by SAS axiom the two triangles are congruent.

Darina [25.2K]1 year ago

5 0

ABCD is a parallelogram Given
AE=CE, BE=DE The diagonals of a parallelogram are bisect each other
∠AEB=∠CED Vertical angles are congruent
ΔABE is congruent to ΔCDE SAS theorem

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1 year ago

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2 years ago

Read 2 more answers

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$