Different Methods to solve a Plane Geometry Question

Different Methods to solve a Plane Geometry Question　

C.M. Chung, Department of Curriculum and Instruction, CUHK

W.M. Chiu, United Christian College

　

Introduction

The converse of a theorem in plane geometry sometimes provides a good exercise for nurturing high-level thinking in students, even though the proof of the original theorem is straightforward. In some cases, the converse is a much more difficult theorem to prove, e.g. the converse of theorem which asserts that an isosceles triangle has two bisectors of the base angles equal. In some cases, the converse of the theorem can be proved in many ways. In this article the authors will illustrate the latter category by the following question.

Question

Given that ABCD is a square, and O is a point inside the square such that Ð ODC = Ð OCD = 15° . Prove that AOB is an equilateral triangle.

(Note that if O is a point inside the square such that AOB is an equilateral triangle, then it is straight-forward to prove Ð ODC = Ð OCD = 15° .)

Method 1

Construct D PAD @ D ODC inside ABCD.
DO = CO (Sides of opp. eq. Ð s)

\ AP = PD = DO = OC

Also Ð PAD = Ð PDA = Ð ODC = 15°

Ð ADC = 90° (Property of a square)

\ Ð PDO = 60°

Ð DPO = Ð DOP (base Ð s of isos. D )

\ Ð DPO = 60° (Ð sum of D )

Ð APD = 150° (Ð sum of D )

\ Ð APO = 150° (Ð s at a point)

Ð PAD = Ð PDA (base Ð s of isos. D )

\ Ð PAD = 15° (Ð sum of D )

Ð DAB = 90° (Property of square)

\ Ð OAB = 60°

Similarly, Ð OBA = 60°

\ Ð AOB = 60° (Ð sum of D )

\ D AOB is equilateral.

Method 2

Ð ADC = Ð BCD = 90° (Property of square)
Ð ADO = Ð BCO = 75°

DO = CO (sides of opp. eq. Ð s)

AD = BC (Property of square)

\ D ADO @ D BCO (SAS)

\ AO = BO

Ð AOD = Ð BOC

Let Ð OAB = x, then Ð OBA = x (Base Ð s of isos D )

Ð AOB = 180° - 2x (Ð sum of D )

Ð DOC = 150° (Ð sum of D )

\ Ð AOD = x + 15° so Ð ADO = 75°

If x < 60° then Ð AOD < Ð ADO = 75° \ AD < AO

Also Ð ABO < Ð AOB \ AO < AB

i.e. AD < AB which contradicts the fact that ABCD is a square.

\ x cannot be smaller than 60° .

Similarly, x cannot be larger than 60°

\ x = 60°

\ Ð ABO = Ð BAO = Ð AOB = 60°

\ D ABC is equilateral.

　

Method 3

Produce CO to meet BD at P. Join AP.
Ð ADP = Ð CDP = 45° (Property of square)

DP = DP (common)

AD = DC (Property of square)

\ D ADP @ D CDP (SAS)

\ AP = PC

Ð DAP = Ð DCP = 15°

Ð DPA = Ð DPC

Ð PDO = 30°

Ð POD = 30° (ext. Ð of D )

\ PO = PD (Sides of opp. eq. Ð s)

Ð DPO = 120° (Ð sum of D )

\ Ð DPA = 120°

Ð APO = 120° (Ð s at a point)

\ D APO @ D CPD (SAS)

\ AO = CD

Similarly BO = CD

\ AO = BO = AB = CD

\ D AOB is equilateral.

Method 4

Construct an equilateral D CDP as shown.
DO = CO (opp. sides of eq. Ð s)

OP = OP (common sides)

DP = CP (sides of equil. D )

\ D DOP @ D COP (SSS)

Ð DPO = Ð CPO = 30°

Ð ADO = 75° = Ð PDO

AD = PD = CD (Sides of square)

DO = DO (common)

\ D ADO @ D PDO (SAS)

\ Ð DAO = Ð DPO = 30°

\ Ð BAO = 60°

Similarly Ð ABO = 60°

\ Ð AOB = 60° (Ð sum of D )

\ D AOB is equilateral.

Method 5

Construct an equilateral D CPD as shown.
Ð PCD = 60° (Ð of an equil. D )

\ Ð BCP = 30° (Property of square)

Ð BPC = Ð PBC = 75° (base Ð s of isos. D )

\ D ABP @ D DOC (ASA)

\ PB = OC (corr. sides)

PO = PO (common)

Ð BPO = Ð COP = 105°

\ D PBO @ D OCP (SAS)

\ OB = CP (corr. sides)

OB = CP = CB = AB

Similarly OA = DP = DA = AB

\ D AOB is equilateral.

Method 6

Applying sine and cosine laws.
Let AB = BC = CD = DA = 1 (Property of square)

\ s₁ =

s² = 1² + s₁² - 2 ´ 1 ´ s₁ ´ cos 75°

= 1 +

= 1 +

= 1 +

= 1

\ s = 1

Similarly AO = 1

\ OB = OA = AB = 1

\ D AOB is equilateral.

Method 7

Construct an equilateral triangle AQB with Q inside the square.

It is straight-forward to prove

Ð QDC = Ð QCD = 15° .

Hence, QD, QC coincide OD , OC respectively, i.e. Q and O coincide.

\ AOB is same as AQB, which is an equilateral triangle.

　

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