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)
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.

Proof: In triangles ABD and ACD:
Angle BDA = Angle CDA = 90° (given)
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° ( sum of triangle angles = 180° ).
Therefore Angle KPM = 80° (LPM is a straight line).
Now, since triangle KPM is isosceles, Angle PKM = Angle PMK
so
Angle PKM=Angle PMK = 50°(sum of triangle angles = 180°).

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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 = ½ (180° 120°)
so Angle DAC = Angle DCA = 30°.
ABE is a straight line, therefore Angle ABC = 70°.
Angle BAC = 180° ( 70° + 40°) = 70°.

This example illustrates the theorem that says:
the exterior angle of a triangle is equal to the sum
of the interior and opposite angles
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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 XYZ 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.

Given: CE and DF are both perpendicular to line AB.
Req'd to Prove: CE // DF
Proof: AB is a transversal cutting CE and DF at C and D.
Angle ECB = Angle FDA = 90° (given)
so Angle ECB + Angle FDA = 180° (addition)
therefore, Angle ECB and Angle FDA are supplementary angles (definition)
so CE // DF since the interior angles are supplementary (parallel line theorem).

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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:

Proof: Since AB // CD, Angle ABD = Angle BDC (alternate angles)
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 XYZ. 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 Angle PYM = Angle PYN (congruence)
therefore PY bisects Angle XYZ by definition.

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