Chapter 23 Geometry Eight More Methods to Draw Auxiliary Lines (1)

50 AMC Lectures Chapter 23 Eight More Methods To Draw Auxiliary Lines

1

BASIC KNOWLEDGE

In this lecture, we will introduce eight commonly used methods to draw auxiliary lines.

1. Double the length of the median of a triangle.

In triangle ABC, AD is the median on side BC. If we extend AD to E

such that DE = AD and connect CE, we get two congruent triangles

CDE and BDA.

Example 1: In ABC, AD is the median. BE and AC meet at E. BE and AD meet at F. If

AE = EF, show that AC = BF.

Proof:

Extend AD to H such that DH = AD.

Since BD = CD and BDH =ADC, then ACD HBD, AC =

BH, and DAC =H.

We are given that AE = EF, so AFE = EAF = BFH.

Therefore in BFH, BFH =H, BF = BH = AC .

2. Draw the height of the figure (especially when area calculation is involved).


2

Example 2: (1972 AMC) A triangle has angles of 30 and 45. If the side opposite the

45 angle has length 8, then the side opposite the 30 angle has length

(A) 4 (B) 24 (C) 34 (D) 64 (E) 6

Solution: (B).

Let s denote the length of the side we want to figure out (see

figure to the right). The altitude to the longest side, opposite

the 30 angle, has length 42

8 and is also one leg of an

isosceles right triangle with hypotenuse s, which therefore has length 24 .

Example 3: (1973 AMC) Two congruent 30 – 60 – 90 triangles are placed so that they

overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the

area common to both triangles is

(A) 36 (B) 38 (C) 39 (D) 312 (E) 24

Solution: (D).

In the adjoining figure MV is an altitude of ∆AMV (a 30 –

60 – 90 triangle), and MV has length 32 . The required

area of triangle ABV, therefore is

1

2(AB)(MV ) 3212

2

1 312 .

3. Draw the diagonals of a parallelogram.

ABCD is a parallelogram. AC and BD are the diagonals. 222222 DACDBCABBDAC

Proof:

Draw DE AB, CF AB.

By the Pythagorean Theorem,

BFABBCABBFBCBFABCFAFAC 2)()( 22222222 (1)

AEABADABAEADAEABDEBEBD 2)()( 22222222 (2)

Since ∆ADE ∆BCF (AD = BC, DE = CF, AED = BFC = 90), AE = BF.


3

(1) + (2): 222222 DACDBCABBDAC .

Theorem: A diagonal of a parallelogram divides the parallelogram into two congruent

triangles.

AED CEB and AEB CED.

Theorem: The diagonals of a parallelogram bisect each other. The

converse, if two diagonals of a quadrilateral bisect each other, the

quadrilateral is a parallelogram, is also true.

AE = EC and DE = EB.

Example 4: In triangle ABC, if AB = c, AC = b, BC = a and O is the

midpoint of AC. Find mb, the length of the median BO.

Solution:

Extend BO to D such that BO = OD. Connect AD and CD. Since AC and

BD bisect each other, they are two diagonals of a parallelogram, and ABCD is a

parallelogram.

Therefore 22222222 4)2

(4)2

(4)2

(4)(2 BOACBDAC

BDACBCAB

])2

([2 2222 ACBOBCAB ])

2([2 2222 b

mac b

222 222

1bcamb .

This is the formula to calculate the median of a triangle if three sides are known.

Similarly, we can have: 222 222

1acbma and 222 22

2

1cbamc .


4

Example 5: DB is the diagonal of parallelogram ABCD. EF//DB and

meets BC at E and DC at F, respectively, as shown in the figure.

Show that triangle ABE and triangle ADF have the same areas.

Solution:

Draw AC, the second diagonal of ABCD.

BC

BE

S

S

ABC

ABE

; DC

DF

S

S

ADC

ADF

.

Since EF ⁄⁄DB, DC

DF

BC

BE

ABC

ABE

S

S

ADC

ADF

S

S

We also know that ADCABC SS (∆ABC ∆CDA).

Therefore ADFABE SS .

4. Translating a diagonal or a leg of trapezoid to form a parallelogram.

(1). In trapezoid ABCD, AB//DC. Draw BE such that

BE//AC and meet the extension of DC at E.

We get:

BE = AC, AB = CE, DE = DC + CE = DC + AB

When AD = BC, we get BD = AC = BE.

(2). In trapezoid ABCD, AB//DC. Draw CF so that CF//AD to

meet AB at F.

We get: AD = FC and AF = DC.

Example 6: (1970 AMC) In the accompanying figure, segments AB and CD are

parallel, the measure of angle D is twice that of angle B, and the

measures of segments AD and CD are a and b respectively. Then

the measure of AB is equal to

(A) ba 22

1 (B) ab

4

3

2

3 (C) 2a – b (D) ab

2

14 (E) a + b.


5

Solution: (E).

Let the bisector of D intersect AB at P (see figure). Then the alternate interior angles

APD and PDC as well as ADP are equal to angle B, so

∆APD is isosceles with equal angles at P and D. This

means that AP = AD = a. Since PBCD is a parallelogram,

we have PB = DC = b; so AB = AP + PB = a + b.

Example 7: In a convex quadrilateral ABCD, AD BC. Show

that ACBD if AC2 + BD

2 = (AD + BC)

2.

Solution: Draw DE so that DE//AC and DE meets the extension

of BC at E. Then ∆CED ∆DAC and DE = AC, CE = AD.

In ∆BDE, BE = BC + CE = BC + AD. DE = AC.

Since AC2 + BD

2 = (AD + BC)

2, or DE

2 + BD

2 = BE

2, by the

converse of the Pythagorean theorem, BDE = 90.

Therefore BDDE. We also know that AC ⁄⁄ DE, so ACBD.

5. Draw the perpendicular to chord through the center of a circle

O is the center of the circle. Draw OC AB.

We have AC = CB and AD = DB.

Theorem: A line perpendicular to a chord of a circle and containing the center of the

circle, bisects the chord and its major and minor arcs.

Theorem: The perpendicular bisector of a chord of a circle contains the center of the

circle.


6

Example 8: (1973 AMC) A chord which is the perpendicular bisector of radius of length

12 in a circle, has length

(A) 33 (B) 27 (C) 36 (D) 312 (E) none of these

Solution: (D).

Let O denote the center of the circle, and let OR and AB be the radius and the

chord which are perpendicular bisectors of each other at M. Applying the

Pythagorean theorem to right triangle OMA yields (AM)2 = (OA)

2 – (OM)

2 =

1221 – 6

2 = 108, AM = 36 .

Thus the required chord has length 312 .

6. Draw the inscribed angle of the diameter

Theorem: An angle inscribed in a semicircle is a right angle.

Theorem: The measure of an inscribed angle equals one-half the measure

of its intercepted arc.

902

180C

Example 9: (1995 AMC) In the figure, AB and CD are diameters of

the circle with center O, AB CD, and chord DF intersects AB at E.

If DE = 6 and EF = 2, then the area of the circle is

(A) 23 (B) 2

47 (C) 24 (D)

2

49 (E) 25

Solution: (C).

Draw segment FC. Angle CFD is a right angle since arc CFD is a

semicircle. Then right triangles DOE and DFC are similar to each

other, so the following equality holds true:

DC

DE

DF

DO .

Let DO = r and DC = 2r. Substituting this into the equality

above, we have

r

r

2

6

8 482 2 r 242 r .


7

The area of the circle is r2 = 24.

Example 10: AB is the diameter of circle O. C is a point on the circumference. P is a

point on the circumference PF is perpendicular to AB. PF meets

AC at E, AB at D and the extension of BC at F. Show

that DFDEDP 2 .

Solution:

Connect PA and PB. AB is the hypotenuse of right triangles APB

and ACB, so APB = 90 and ACB = 90.

Since triangle APB is a right triangle, DBADDP 2 . Instead of

showing that DFDEDP 2 , we can now prove that DFDEDBAD .

Note that ∆ADE~∆FDB.

We know that F = EAD, and ADE =FDB = 90.

Therefore

∆ADE~∆FDB DB

DF

DE

AD DFDEDBAD .

7. When two circles intersect, draw the common chords or connect the centers.

O1O2 is the perpendicular bisector of EF. O2C DC.

Theorem: Any point on the perpendicular bisector of a line

segment is equidistant from the endpoints of the line

segment. Two points equidistant from the endpoints of a

line segment determine the perpendicular bisector of the

line segment.


8

Example 11: (1966 AMC) The length of the common chord of two intersecting circles is

16 feet. If the radii are 10 feet and 17 feet, a possible value for the distance between the

centers of the circles, expressed in feet, is:

(A) 27 (B) 21 (C) 389 (D) 15 (E) undetermined

Solution: (B).

Denote the common chord by AB, its midpoint by P, and the

centers of the smaller and larger circles by O and O; OO is

perpendicular to AB and passes through P.

The Pythagorean Theorem applied to right triangles OPA and

OPA yields

6, ,36810 22222 OPAPOAOP

And

O P 2 O A2 AP2 172 82 225, O P 15.

The distance between the centers of the circles is 15 + 6 = 21.

8. When a figure looks like a part of the other figure, draw the original figure.

Example 12: (1968 AMC) Let side AD of convex quadrilateral ABCD be extended

through D, and let side BC be extended through C, to meet in point E. let S represent the

degree-sum of angles CDE and DCE, and let S represent the degree-sum of angles BAD

and ABC. If r = S

S

, then:

(A) r = 1 sometimes, r > 1 sometimes (B) r = 1 sometimes, r < 1 sometimes

(C) 0 < r < 1 (D) r > 1 (E) r = 1

Solution: (E).

Extended AD and BC to meet at E, and form the original figure,

triangle ABE.

We know that the sum of the angles in a triangle is 180(see diagram

on right), so

E + CDE +DCE = E + S = 180 in ∆EDC

and


9

E + BAD + ABC = E + S = 180 in ∆EAB.

Hence S = S = 180 – E, so r = S

S

= 1.

Example 13: (1972 AMC) Let ABCD be a trapezoid with the measure of base AB twice

that of base DC, and let E be the point of intersection of the diagonals. If the measure of

diagonal AC is 11, then that of segment EC is equal to

(A) 3

23 (B)

4

33 (C) 4 (D)

2

13 (E) 3

Solution: (A).

Extend sides AD and BC to meet at V, forming the original

figure, triangle ABV.

In ∆ABV, AC and BD are medians from vertices A and B that

meet at point E, which divides the length of each in the ratio 2

: 1. This means that

.3

23

3

11

3

1 ACEC

Example 14: (1999 AMC #23) The equiangular convex hexagon ABCDEF has AB = 1,

BC = 4, CD = 2, and DE = 4. The area of the hexagon is

(A) 32

15 (B) 39 (C)16 (D) 3

4

39 (E) 3

4

39

Solution:

Extend FA and CB to meet at X, and BC and ED to meet at Y, and

DE and AF to meet at Z. Triangle XYZ is the original figure. The

interior angles of the hexagon are 120, so triangles XYZ, ABX, CDY,

and EFZ are all equilateral. Therefore, because AB = 1, BX = 1; since

CD = 2, CY = 2. Thus XY = 7 and YZ = 7. Since YD = 2 and DE = 4,

EZ = 1. The area of the hexagon can be found by subtracting the areas

of the three small triangles from the area of the large triangle:

4

343)

4

3(1)

4

3(2)

4

3(1)

4

3(7 2222 .


10

PROBLEMS

Problem 1. (1967 AMC) In quadrilateral ABCD with diagonals AC and BD intersecting

at O, BO = 4, OD = 6, AO = 8, OC = 3, and AB = 6. The length

of AD is

(A) 9 (B) 10 (C) 36 (D) 28 (E) 166

Problem 2: (2003 AMC 12 B) In rectangle ABCD, AB = 5 and BC = 3. Points F and G

are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E.

Find the area of AEB.

(A) 10 (B) 2

21 (C) 12 (D)

2

25 (E) 15

Problem 3: Rectangle ABCD has the length a and width b. M is the midpoint of BC. DE

AM at E. Prove: .4

222 ba

abDE

Problem 4: Hypotenuse AB of right ABC is divided into four congruent segments by

points G, E, and H, in the order A, G, E, H, and B. If AB = 20,

find the sum of the squares of the measures of the line segments

from C to G, E, and H.


11

Problem 5: (AMC) In the figure, ABCD is an isosceles trapezoid with side lengths AD =

BC = 5, 4AB , and DC = 10. The point C is on DF and B

is the midpoint of hypotenuse DE in the right triangle DEF.

Then CF =

(A) 3.25 (B) 3.5 (C) 3.75 (D) 4.0

(E) 4.25

Problem 6: (1984 AMC) In ∆ABC, D is on AC and F is on BC. Also, ABAC, AFBC,

and BD = DC = FC =1. Find AC

(A) 2 (B) 3 (C) 3 2 (D) 3 3 (E) 4 3

Problem 7: (1975 AMC) In the adjoining figure triangle ABC is such that AB = 4 and AC

= 8. If M is the midpoint of BC and AM = 3, what is the length of BC?

(A) 262 (B) 312 (C) 9 (D) 1324

(E) not enough information given to solve the problem

Problem 8. In ABC, AB = AC. E is the midpoint of AB. Extend AB to D such that BD =

BA. Prove: CD = 2CE.


12

Problem 9: (1972 AMC) Inside square ABCD (see figure) with sides of length 12 inches,

segment AE is drawn, where E is the point on DC which is 5 inches from

D. The perpendicular bisector of AE is drawn and intersects AE, AD, and

BC at points M, P, and Q respectively. The ratio of segment PM to MQ is

(A) 5:12 (B) 5:13 (C) 5:19 (D) 1:4 (E) 5:21

Problem 10: (1959 AMC) In triangle ABC, BD is a median. CF intersects BD at E so BE

= ED. Point F is on AB. Then, if BF = 5, BA equals:

(A) 10 (B) 12 (C) 15 (D) 20 (E) none of these

Problem 11: (1959 AMC) The base of a triangle is 80, and one of the base angles is 60.

The sum of the lengths of the other two sides is 90. The shortest side is

(A) 45 (B) 40 (C) 36 (D)17 (E)12

Problem 12: In square ABCD, E, F, G, H are points in each side. EG HF. Prove: EG =

HF.

Problem 13: In ∆ABC, AC = 3AB, CDAO at D. AO is the angle bisector of BAC.

Show that AO = DO.


13

Problem 14: (1987 North Carolina Math League) A trapezoid is inscribed in a semicircle

of radius r = 5 with the diameter of the semicircle serving as one of the bases of the

trapezoid. Let h be the height of the trapezoid. Determine the value of h for which the

area of the trapezoid is as large as possible.


14

SOLUTIONS:

Problem 1. Solution: (E).

Denote F as the foot of the perpendicular from A to an

extended diagonal DB, and denote BF and FA by x and y

respectively (see figure). By the Pythagorean Theorem, 222 6 yx and 222 8)4( yx .

Subtracting the first of these equations from the second

yields

8x + 16 = 28, x = 3/2.

Substitute the value of x into the first equation and solve for

y2:

222 6)2

3( y

4

1352 y .

Therefore 1664

664

4

135)

2

23()10( 2222 yxAD .

166AD .

Problem 2: Solution: (D).

Let H be the foot of the perpendicular from E to DC . Since CD = AB = 5,

it follows that FG = 2. Because ∆FEG is similar to ∆AEB, we have

,5

2

3

EH

EHso 5EH = 2EH + 6, and EH = 2.

Hence the area of ∆AEB is

.2

255)32)(

2

1(

Problem 3: Solution: .4

222 ba

abDE

Connect DM.

By the Pythagorean Theorem,

2222 42

1baBMABAM .


15

Since ABCDADM SS2

1 ,

1

2AM DE

1

2ab.

Substituting the value of AM into the equation above and solving for DE:

.4

222 ba

abDE

Problem 4: Solution: 350.

Draw CJ perpendicular to AB at J. Since AB = 20, CE = 10.

Let GJ = x, and JE = 5 – x.

By the Pythagorean Theorem, in right triangles ∆CJG and ∆CJE,

(CG)2 – x

2 = 10 – (5 – x)

2

or (CG)2 = 75 +10x. (1)

Similarly, in ∆CJH and ∆CJE, 222 )5()10()( xxCH ,

or (CH)2 = 175 – 10x. (2)

By the addition of (1) and (2):

.350100101751075)()()( 222 xxCECHCG

Problem 5: Solution: 4.

Method 1 (Official Method)

(D) Drop perpendiculars AG and BH to DF . Then GH =

4, so 3)(2

1 GHDCHCDG

Since BH ⁄⁄ EF and B is in midpoint of DE, it follows that H

is the midpoint of DF. Thus,

DH = DG + GH = 3 + 4 and DF = 2DH = 14, so CF = DF – DC = 14 – 10 = 4.

Method 2 (our solution)

Drop perpendiculars AG and BH to DF.


16

Connect BF. Then GH = 4, so 3)(2

1 GHDCHCDG . Since BD = BF, triangle

DBF is isosceles, so DH = HF and CF = 4.

Problem 6: Solution: (C)

Draw DG so that DGBC and G lies on BC. Let AC = x and

GC = y. Note that BC = 2y, since ∆BCD is isosceles.

Since ∆DCG ~ ∆ACF ~ ∆BCA, we obtain the equal ratios:

.2

1

1

x

yx

y

Thus x

y1

and ,2

2xy implying that

x3 = 2, or 3 2x .

Problem 7: Solution: (B).

Recall that the sum of the squares of the sides of a parallelogram is equal to the sum of

the squares of its diagonals. Applying this to the parallelogram having AB and AC as

adjacent sides yields

AD2 BC2 AB2 CD2 AC2 BD2 2222 6)84(2 BC , 1246)84(2 2222 BC 312BC .

Problem 8. Proof :

Method 1:

Find the midpoint of CD, F. Connect BF to get DF = CF.

Since AB = BD, so BF=1/2 AC. Since BF//AC, so 1=ACB.

Because AB = AC, BE = AE =1/2 AB, so ACB =2, BE = BF and 1

=2.

Since BC = BC, BEC BFC.

Therefore EC = CF, and so CD = 2CF = 2CE, i.e. CD = 2CE.

Method 2:


17

Extend CE to F such that CE = EF. Since AE=EB and 1=2, AEC

BEF

Thus, 3 = F, 4 = A, BF = AC. Since AB = AC = BD, BF = BD.

DBC = A + ACB = A + ABC or FBC = 4 + ABC = A

+ABC .

So DBC = FBC. Since BC = BC, FBC DBC.

Therefore CF = CD.

Since CE = EF = 1/ CF, CD = 2CE.

Method 3:

Extend AC to F such that CF = AC.

So AF = AD. AC = AB, A = A, ACD ABF.

So CD = BE.

Since AE = BE, CA = CF, EC = 1/2 BF, BF = 2CE, CD

= 2CE.

Method 4:

Extend BC to F such that CF = BC.

Since AB=AC, ABC=ACB, DBC = ACF.

Since AC = AB, AB = BD, BD = AC,

ACF DBC, CD = AF.

Since AE = EB, BC = CF, so AF = 2CE, i.e. CD =

2CE.

Method 5:

Find F, the midpoint of AC. Connect BF, then CE = BF.

Since AB = BD and AF = CF, 2BF = CD. Therefore CD =

2CE.


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Method 6:

Extend CE to F so that CE = EF.

It is easy to prove that AEF BEC (SAS).

So AF = BC, FAE = ABC, and AC = BD. We have AF//BC.

Therefore FAC + ABC = 180.

Since ABC + DBC = 180, then DBC = FAC and

FAC CBD, CF = CD.

Since CF = 2CE, so CD = 2CE.

Problem 9: Solution: (C).

Let the line through M parallel to side AB of the square intersect sides AD

and BC in points R and S, respectively. Since M is the midpoint of AE,

2

5

2

1 DERM inches , and therefore

2

19

2

512 MS inches.

Since PMR and QMS are similar right triangles, the required ratio PM:MQ

= RM:MS = 5:19 because corresponding sides of similar triangles are proportional.

Problem 10: Solution: (C).

Let G be a point on EC so that FE = EG. Connect D with G, forming

parallelogram FDGB.

DG = 5, AF = 10, AB = 15.

Problem 11: Solution: (D).

Let the triangle be ABC, with AB = 80, BC = a, CA = b = 90 – a, B

= 60. Let CD be the altitude to AB, and let x = BD.

xCD 3 , a = 2x, b = 90 – 2x;

3x2 + (80 – x)

2 = (90 – 2x)

2;

2

17x ; and a = 17, b = 73.


19

Problem 12: Solution:

In order to prove EG = HF, we should try to construct two congruent triangles with

corresponding sides of EG and HF.

First, we move AB to the position of HN, and BC to the position of EM.

GEM = FHN, GME = FNH = 90, and EM = BC = AB = HN.

Therefore GEM FHN and EG = HF.


Since 1 = 2 and ADCD, we can construct an isosceles triangle ACE by extending

AB and CD to meet at E.

Therefore AE = AC = AB + BE = AB + 2AB

Connect DF such that F is the midpoint of BE.

Since F is the midpoint of BE and D is the midpoint of CE, FD is

the midline of triangle EBC, and therefore FD//BC, or BO//FD.

We know that AB = BF and B is the midpoint of AF.

Therefore BO divides AD into two equal parts. That is, AO = OD.


5 3

2.

If we reflect the trapezoid about the diameter, we will form a hexagon inscribed in the

circle. In order for the area of the trapezoid to be as large as possible, the area of the

hexagon should be maximized. The hexagon with the largest area will be a regular

hexagon whose sides are equal to the radius. Thus the height of the trapezoid is h =

3

2r

5 3

2.

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Chapter 23 Geometry Eight More Methods to Draw Auxiliary Lines (1)