Proving - Congruent Triangles

Category: Plane Geometry

"Published in Newark, California, USA"

Given the figure below:

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1. Given AF ≅ AD, and FE ≅ ED. Prove that ΔAFE ≅ ΔAED.

2. Given ABDE is a square and ΔACD is an isosceles triangle.  Prove that ΔAED ≅ ΔBDC.

3. Given ABDE is a square. Prove that ΔABD ≅ ΔADE.

4. Given AB ≅ DE, AD bisects ∠BAE and ∠BDE. Prove that ΔBAD ≅ ΔADE.

5. Given AF ≅ DC and DC ≅ ED. Prove that ΔAFD ≅ ΔACD.

Solution:

Consider Case 1:

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Proof:

  

1. Statement: AF ≅ AD and FE ≅ ED.

    Reason: Given items.

2. Statement: AE ≅ AE. 

    Reason: Reflexive property of congruence.

Therefore, ΔAFE ≅ ΔAED.

Reason: SSS (Side-Side-Side) Postulate

Consider Case 2:

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Proof:

1. Statement: ABDE is a square and ΔACD is an isosceles triangle.

    Reason: Given items.

2. Statement: AE ≅ BD ≅ AB ≅ ED

    Reason: All sides of a square are congruent.

3. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

4. Statement: ∠EDA ≅ ∠BAD

    Reason: The alternating interior angles of a two parallel lines that passes a transversal line are congruent. The two opposite sides of a square are parallel.

5. Statement: ∠DAB ≅ ∠BCD

    Reason: The base angles of an isosceles triangle are congruent.

6. Statement: AD ≅ DC

    Reason: The two sides of an isosceles triangle are congruent.

7. Statement: AB ≅ BC

    Reason: The base altitude (BD) of an isosceles triangle bisects the line segment (AC) of a base.

Therefore, ΔAED ≅ ΔBDC.

Reason: SAS (Side-Angle-Side) Postulate

Consider Case 3:

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Proof:

1. Statement: ABDE is a square.

    Reason: Given item.

2. Statement: AE ≅ BD ≅ AB ≅ ED

    Reason: All sides of a square are congruent.

3. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

Therefore, ΔABD ≅ ΔAED.

Reason: SSS (Side-Side-Side) Postulate

Consider Case 4:

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Proof:

1. Statement: AB ≅ ED, ∠EAD ≅ ∠BAD, and ∠BDA ≅ ∠ADE.

    Reason: Given items.

2. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

Therefore, ΔBAD ≅ ΔADE.

Reason, ASA (Angle-Side-Angle) Postulate.

Consider Case 5:

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Proof:

1. Statement: AF ≅ DC and DC ≅ ED.

    Reason: Given items.

2. Statement: AB ≅ ED and AE ≅ BD.

    Reason: The opposite sides of a rectangle are congruent.

3. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

4. Statement: AE ┴ ED and BD ┴ AB.

    Reason: The sides of a rectangle ABDE are perpendicular to each other.

5. Statement: ∠FEA ≅ ∠AED, and ∠ABD ≅ ∠DBC.

    Reason: The sum of the supplementary angles is 180. If ∠AED and ∠ABD are 90°, then ∠FEA and ∠DBC msut be 90°.

6. Statement: ΔFEA and ΔDBC are right triangles.

    Reason: One of the angles of each triangles is 90°.

7. Statement: FE ≅ BC

    Reason: Since ΔFEA and ΔDBC are right triangles, we can use Pythagorean Theorem (c² = a² + b²) to solve the other side of a right triangle. If AF ≅ CD and AE ≅ BD, then FE ≅ BC. 

8. Statement: FD ≅ AC

    Reason: Since FE ≅ BC and ED ≅ AB, then FE + ED ≅ AB + BC. 

Therefore, ΔAFD ≅ ΔACD.

Reason: SSS (Side-Side-Side) Postulate

Volume Derivation - Sphere

Category: Integral Calculus, Solid Geometry

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Given the equation of the sphere below:

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Prove that the volume of sphere is  ⁴/₃ π r³                                                      

Solution:

From the given figure, let's consider only the ¹/₈ of the sphere to get the volume and then multiply it by 8 later. 

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Image that there are many tiny cubes that will fill-up the section of a sphere. Let's assume that the dimensions of the cubes, dx, dy, and dz are perfect enough to fill-up the section of a sphere. 

dV = dx dy dz

Integrate on both sides of the equation, we have

Next, let's assign the lower and upper limits for each integrals.

Along the x-axis, let's consider the largest cross section of a sphere at xy plane, the cube is moved from 0 to r, and so the lower limit is 0 and the upper limit is r.

Along the y-axis, let's consider the largest cross section of a sphere at xy plane. The cube is moved from 0 to the surface of the cross section. The cross section is a circle with an equation: x² + y² = r² where z² = 0. Solve for the value of y, we have

Therefore, the lower limit is 0 and the upper limit is 

Along the z-axis, let's consider the largest cross section of a sphere at xy plane. The cube is moved from the cross section of a sphere at xy plane to the surface of a sphere. From the equation of the sphere: x² + y² + z² = r², solve for the value of z, we have

Therefore, the lower limit is 0 and the upper limit is

The volume of a sphere can be written as

The above equation can also be written as

Integrate first with respect to z, consider x and y are the constants

   let a² = r²- x²

   and u² = y²

Using the integration formula,

Integrate with respect to y, consider x as a constant. Therefore,

Finally, integrate with respect to x, we have

 

Therefore, the volume of a sphere is 

Proving Trigonometric Identities - Double Angle

Category: Trigonometry

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Prove the trigonometric identity for

Solution:

In proving of trigonometric identities, you must choose the more complicated side of the equation and then simplify as much as you can until it is equal to the other side of the equation. In this case, consider the given equation

Let's expand and then simplify the left side of the equation as follows

but 

The equation becomes

Therefore,