Geometry: Proving Deductions, Examples |

**Proving Deductions**

When we need to prove a statement **(deduction**) in geometry, we must proceed in a specifically defined manner; that is, we must present the deduction in a very precise way. Every deduction we do must have a **diagram**, a **given** statement, a **required to prove** statement, and a **proof**.** **The** proof** consists of statements which are based on **given** facts, **axioms** and **theorems**, and **each statement** in the proof **must be justified. **The final statement of your proof must be exactly the same as the statement in the** required to prove**.

Notice in the proofs for the triangle congruence theorems in lesson 3.1, each statement is justified by a statement that is **given** or some other authority such as an **axiom** or **proposition** (theorem).

Notice also that **all proofs begin with a given statement or a construction** because sometimes we have to add a line or other shape to the diagram to facilitate the proof.

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Let's prove the **isosceles triangle theorem** as a deduction now that

we can prove triangles are congruent.

**Given**: Triangle ABC in which AB = AC.

**Req'd to Prove**: ANGLE B = Angle C

**Proof**: Construct AD, the bisector of Angle A to meet BC at D.

- In triangles ABD and ACD: AB = AC (given)

Angle BAD = Angle CAD (definition of angle bisector)

AD = AD (common side or reflexive property)

Therefore (Side, Angle, Side)

So Angle B = Angle C (definition of congruence).

Now let's do a number of deductions using the theorems and axioms we've learned.

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**Example 2**

**Given:** Quadrilateral ABCD in which AB = BC and AD = DC.

**Req'd to Prove:**

- 1) Angle A = Angle C

2) If we join B to D, then .

**Proof 1:**

- Join A to C

In triangle ABC, Angle BAC = Angle BCA (isosceles triangle thm.)

In triangle ADC, Angle DAC = Angle DCA (isosceles triangle thm.)

So Angle BAC + Angle DAC = Angle BCA + Angle DCA ( equal sums axiom)

Therefore Angle A = Angle C.

**Proof 2:**

Join B to D

In triangle ABD and triangle BCD:

AB = BC and AD = CD (given)

BD = BD (common side)

So (SSS)

We have now proven both statements in the required to prove.

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**Example 3**

**Given:** Quadrilateral ABCD in which AB is both equal and parallel to CD.

**Req'd to Prove:** The diagonal BD divides the quadrilateral into 2 congruent triangles.

**Proof:**

- BD is a transversal for the parallel lines AB and CD.

In triangles ABD and CBD:

AB = CD (given)

Angle ABD = Angle BDC (alternate angles)

BD = BD (common side)

So (SAS)

**Example 4**

**Given:** Triangle ABC in which AD, the bisector of Angle A is perpendicular to BC at D.

**Req'd to Prove:** triangle ABC is isosceles.

Angle BAD = Angle CAD (definition of bisector)

Angle BDA = Angle CDA = 90° (given)

AD = AD (common side)

So (AAS)

Therefore AB = AC (definition of congruence)

So triangle ABC is isosceles. (definition of isosceles)

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**Numerical Exercises** are questions in which we must find numerical values for lengths or angle measures. In such exercises, we do not have to follow the same formal approach as we do when proving deductions. It helps however to justify our statements so we understand our reasons for making those statements.

**Example 5:** In triangle ABC, find the measures of angles ABD, DBC, and ACB.

Since the three angles of a triangle add to 180° ,

Angle ACB = 180° – (90° + 35° ) = 55°

Similarly, Angle DBC = 180° – (90° + 55° ) = 35° and Angle ABD = 55°

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**Example 6:** Find the measure of all angles in the diagram given that

KP = LP = PM and Angle KLM = 40°.

- Since triangle KPL is isosceles, Angle LKP = 40

So Angle LPK = 100

Therefore Angle KPM = 80

Now, since triangle KPM is isosceles, Angle PKM = Angle PMK

so Angle PKM=Angle PMK = 50

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**Example 7:**

Given the information in the diagram, find the measures of all unmarked angles.

- triangle ADC is isosceles so Angle DAC = Angle DCA =

so Angle DAC = Angle DCA = 30

ABE is a straight line, therefore Angle ABC = 70

Angle BAC = 180

This example illustrates the **theorem** that says:

**the exterior angle of a triangle is equal to the sum
of the interior and opposite angles**.

Note that Angle EBC = Angle ACB + Angle CAB, since the sum of the three angles in the triangle is 180^{°}** **and** **so is Angle EBC + Angle CBA.

So we know that Angle EBC + Angle CBA = 180** ^{°}** = Angle ACB + Angle CAB + Angle CBA.

Now subtract Angle CBA from both sides of the equation to get

Angle EBC = Angle ACB + Angle CAB.

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**Practice**

**A/ Solve these numerical exercises for the measure of all angles not indicated.**

(solution 1) | (solution 2) | (solution 3) | (solution 4) |

**B/ Prove these deductions. Draw a diagram.
Write a Given, Req'd to Prove, and Proof for each question.
Justify every statement of the Proof.**

1) Prove that two lines which are both perpendicular to the same line are parallel to each other.

2) If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

**A median is a line drawn from a vertex of a triangle**

**to the midpoint of the opposite side.**

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3) The median **BD** of **triangle**** ABC** is produced to **E** so that **DE = DB**. Prove that **AE ****// ****BC**.

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4) **ABCD** is a quadrilateral in which **AB ** is both equal and parallel to **CD**.

Prove that the diagonal **BD** divides the quadrilateral into two congruent triangles.

5) **P** is a point within **triangle ****X****YZ** such that the perpendiculars from

**P** to **XY **and **YZ** are equal to each other. Prove that **PY** bisects **Angle ****XYZ**.

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**Solutions:**

**A/ Solve these numerical exercises for the measure of all angles not indicated.**

[Note: to find Angle GED and** **Angle GDE** **we take **½**(180**° ****–** 166**°)
**since 2 angles are = and the sum of 3 angles in the triangle = 180

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**B/ **

1) Prove that two lines which are both perpendicular to the same line are parallel to each other.

Angle ECB = Angle FDA = 90

so Angle ECB + Angle FDA = 180

therefore, Angle ECB and Angle FDA are supplementary angles (definition)

so

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2) If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

**Given:** ABCD is a quadrilateral in which the diagonals, AC and BD bisect each other at E.

**Req'd to Prove:** that ABCD is a parallelogram.

**Proof:** In triangles AED and BEC:

AE = CE and DE = BE (given)

Angle AED = Angle BEC (opposite angles) so (SAS)

Therefore, Angle EAD = Angle ECB (congruence) so AD // BC (alternate angles)

Similarly, (SAS) so AB // CD (alternate angles)

Therefore, ABCD is a parallelogram by definition.

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3) The median **BD** of **triangle**** ABC** is produced to **E** so that **DE = DB**. Prove that **AE**** // ****BC**.

**Given:** triangle ABC in which BD, the median from B to AC extends to E with DE** **= DB.

**Req'd to Prove:** that AE // BC.

**Proof:** In triangle ADE and triangle BDC:

- AD = DC and ED = DB (given)

Angle ADE = Angle BDC (opposite angles)

(Side, Angle, Side)

therefore Angle AED = Angle DBC (property of congruence)

therefore alternate angles are equal (def. of alternate angles)

therefore AE // BC. (parallel line thm.)

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4) In quadrilateral **ABCD**, **AB ** is **equal and parallel** to **CD**.

Prove that diagonal **BD** divides the **ABCD** into two congruent triangles.

**Given:** Quadrilateral ABCD in which AB = CD and AB // CD.

**Req'd to Prove:**

In triangle ABD and triangle BDC: AB = CD (given)

Angle ABD = Angle BDC (proof)

BD = BD (common side)

(SAS)

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5) **P** is a point within **triangle**** X****YZ. ** **PM** and **PM**, the perpendiculars from

**P** to **XY **and **YZ** are equal to each other.

Prove that **PY** bisects **Angle ****XYZ**. (show that **Angle ****MYP** = **Angle ****NYP**)

**Given:** P is a point within **triangle **XYZ.

PN is perpendicular to YZ and PM is perpendicular to XY. Also, PN = PM.

**Req'd to Prove:** PY bisects **Angle **XYZ.

**Proof:** In the right-angled triangles **PMY** and **PNY**:

- PN = PM (given)

PY = PY (common side)

(hypotenuse and one side)

therefore

therefore PY bisects

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