50 AMC LECTURES Lecture 19 Eight More Methods To ... of AC. Find m b, the length of the median BO. Solution: Extending BO to D such that BO = OD. Connect AD and CD. Since AC and BD

50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines

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BASIC KNOWLEDGE

In this lecture, we introduce eight more 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 AC = BF.

Proof:

Extending AD to H such that DH = AD.

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

BH, 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).



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 required side (see figure).

Then the altitude to the longest side, opposite the 30 angle,

has length 42

8 and is 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 is, therefore, area ))((2

1MVABABV

32122

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.

(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. Also

converse.

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:

Extending 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, that is, 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 .



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//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//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.



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 that ∆APD

is isosceles with equal angles at P and D. This makes 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//AC and meets the extension of BC at E. Then

∆CED ∆DAC. 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 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.



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, so

DC

DE

DF

DO .

Let DO = r and DC = 2r. Substituting, we have

r

r

2

6

8 482 2 r 242 r .

Then 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. APB = 90. ACB = 90.

We have: DBADDP 2 . Now we only need to prove

DFDEDBAD .

We also see that .∆ADE~∆FDB.

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

Therefore ∆ADE~∆FDB DB

DF

DE

AD DFDEDBAD .

7. When two circles are intersecting, 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.

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 chore 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 now yields

6, ,36810 22222 OPAPOAOP

And 15. ,225817 22222 POAPAOPO 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).

Since we extended AD and BC to meet at E, Triangle ABE is the

original figure.

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 E + BAD + ABC = E + S = 180 in ∆EAB.

Hence S = S = 180 – E, so that 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. Triangle ABV is the

original figure.

Then AC and BD are medians from vertices A and B of ∆ABV

meeting 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, 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. Thus the triangles XYZ, ABX, CDY,

and EFZ are equilateral. Since 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 .



PROBLEMS

Problem 1. (1980 AMC) Sides AB, BC, CD and DA of convex quadrilateral ABCD have

lengths 3, 4, 12, and 13, respectively; and CBA is a right angle. The area of the

quadrilateral is

(A) 32 (B) 36 (C) 39 (D) 42 (E) 48

Problem 2. (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 3. In ABC, C=90. D is the middle point of BC.

DEAB at E. Prove ABBEBC 22

Problem 4: (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 5: (1998 AMC) In quadrilateral ABCD, it is given that A = 120, angles B and

D are right angles, AB = 13, and AD = 46. Then AC =

(A) 60 (B) 62 (C) 64 (D) 65 (E) 72

Problem 6: Rectangle ABCD has the length a and width b. M is the middle point of BC.

DE AM at E. Prove: .4

222 ba

abDE

Problem 7: On sides AB and DC of rectangle ABCD, points F and E are chosen so that

AFCE is a rhombus, as shown in the figure. If AB = 16 and BC = 12,

find EF.

Problem 8. On sides AB and DC of rectangle ABCD, points F and E are chosen so that

AFCE is a rhombus, as shown in the figure. If AB = a and BC = b,

find EF.

Problem 9: If the measures of two sides and the included angle

of a triangle are 7, 50 , and 135°, respectively, find the

measure of the segment joining the midpoints of the two given

sides.



Problem 10: 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.

Problem 11: Two circles intersect in A and B, and the measure of the common chord AB

= 10. The line joining the centers cuts the circles in P and Q. If

PQ = 3 and the measure of the radius of one circle is 13, find the

radius of the other circle. (Note that the illustration is not drawn

to scale.)

Problem 12: (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 13: (1961 AMC) In triangle ABC the ratio AC : CB is 3:4. The bisector of the

exterior angle at C intersects BA extended at P (A is between P and B). The ratio PA: AB

is :

(A) 1:3 (B) 3:4 (C) 4:3 (D) 3:1 (E) 7:1

Problem 14: (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 15: (1960 AMC) Given right triangle ABC with legs BC = 3, AC = 4. Find the

length of the shorter angle trisector from C to the hypotenuse:

(A) 13

24332 (B)

13

9312 (C) 836 (D)

6

105 (E)

12

25

Problem 16: (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 17: (1985 AMC) In a circle with center O, AD is a diameter,

ABC is a chord, BO = 5 and ABO = CD = 60. The length of BC is

(A) 3 (B) 33 (C) 2

35 (D) 5 (E) none of the above

Problem 18. In ABC, AB = AC. E is the midpoint of AB. Extend AB to D such that BD

= BA. Prove: CD = 2CE.

Problem 19. In the figure, AC = BD, AD AC, BD BC. Prove AD

= BC.



Problem 20: (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 21: As shown in the figure, AC and BD are two diagonals of trapezoid ABCD

and ACBD. Show that AC2 + BD

2 = (AB + DC)

2 .

Problem 22: (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 23: In quadrilateral ABCD, B = D = 90, A = 60, AB = 4, and AD = 5.

Find BC : CD.

Problem 24: As shown in the figure, ABCD is a trapezoid. Half circle O is inscribed into

ABCD. Find BC if AB = 2 and CD = 3.



Problem 25: (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 26: In square ABCD, E, F, G, H are points in each side. EG HF. Prove: EG =

HF.

Problem 27: As shown in the figure, in equilateral triangle ABC, we extend BA to E and

BC to D such that AE = BD. Connect CD. Show that CE = DE.

Problem 28: In ∆ABC, AC = 3AB, CDAO at D. AO is the angle bisector of BAC.

Show that AO = DO.

Problem 29: A convex octagon ABCDEFG has eight equal interior angles. The lengths

of the sides are 7, 4, 2, 5, 6, 2, x, and y, as shown in the figure. Find its perimeter.



Problem 30: (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.



SOLUTIONS:

Problem 1. Solution: (B).

By the Pythagorean Theorem, diagonal AC has length 5. Since 222 13125 , ∆DAC is a right triangle by the converse of the

Pythagorean theorem. The area of ABCD is

.36)12)(5)(2

1()4)(3)(

2

1(

Problem 2. Solution: (E).

Let F denote the foot of the perpendicular from A to

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

respectively (see figure). Then 222 6 yx and

222 8)4( yx . Subtracting the first of these equations

from the second yields

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

We put 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 3.

Proof:

Method 1:

Since DEAB, ACB =DEB = 90.

B =B, so ABC DBE.

So BD

AB

BE

BC ,

Since D is the middle point of BC, so BD=1/2 BC.

So BC

AB

BE

BC 2 . That is ABBEBC 22



Method 2: (our method)

Draw CGAB. Since CG//DE, and DE cuts BC into two equal

parts, DE also cuts GE into two equal parts, that is GE = EB and

GB = 2BE

AGABAC 2 222 BCACAB (Pythagorean Theorem)

Then ABBEABGBAGABABAGABABBC 2)(22

Problem 4: 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 5: Solution: (B).

Extend DA through A and CB through B and

denote the intersection by E. Triangle ABE is a

30-60-90 triangle with AB = 13, so AE = 26.

Triangle CDE is also a 30-60-90 triangle, from

which it follows that CD = (46 + 26)/ 3 = 24 3 .

Now apply the Pythagorean Theorem to triangle

CDA to find that AC = 62)324(46 22 .

Problem 6: Solution:

Connect DM. 2222 42

1baBMABAM .

Since ABCDADM SS2

1

abDEAM2

1

2

1



Substituting the value of AM and solving for DE: .4

222 ba

abDE

Problem 7:

Method 1

Let AF = FC = EC = AE = x.

Since AF = x and AB = 16, BF = 16 – x.

Since BC = 12, in right ∆FBC, (FB)2 + (BC)

2 = (FC)

2 or (16 – x)

2 + (12)

2 = x

2, and

.2

25x

Again by applying the Pythagorean Theorem to ∆ABC, we get AC = 20.

Since the diagonals of a rhombus are perpendicular and bisect each

other, ∆EGC is a right triangle, and GC = 10.Once more applying

the Pythagorean Theorem in ∆EGC, (EG)2

+ (GC)2 = (EC)

2.

4

625100)( 2 EG , and

2

15EG Thus, FE = 2(EG) = 15.

Method 2:

Since 2

25x (see Method 1), .

2

25EC

Draw a line through B parallel to EF meeting DC at H.

Since quadrilateral BFEH is a parallelogram and FB

= ,2

7 AFAB .

2

7EH Therefore, HC = 9.

In right ∆BCH, 222 )()()( HCBCBH , so BH = 15. Therefore, EF = BH = 15.

Problem 8. Answer: .22 baa

bEF


Draw altitude CD. Since CAB = 135, DAC = 45, therefore,

ADC is an isosceles right triangle. If AC = 50 = 5 2 , then

DA = DC = 5.

In DBC, since DB = 12 and DC = 5, BC = 13 .

Therefore, EF = 2

13

2

1BC .




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

Let GJ = x, and JE = 5 – x. In ∆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 addition of (1) and (2),

.350100101751075)()()( 222 xxCECHCG


Since OA = OB and OA = OB, OO is the perpendicular bisector of AB. Therefore, in

right ∆ATO, since AO = 13 and AT = 5, we find OT = 12. Since OQ = 13 (also a radius of

circle O), and OT = 12, TQ = 1. We Know that PQ = 3. PT = PQ – TQ; therefore, PT = 2.

Let OA = OP = r, and PT = 2, TO = r – 2.

Applying the Pythagorean Theorem in tight ∆ATO, (AT)2 +

(TO)2 = (AO)2

.

Substituting, 52 + (r – 2)

2 = r

2, and

4

29r . PT = PQ + TQ;

therefore, PT = 4. Again, let OA = OP = r then TO = r – 4.

Applying the Pythagorean Theorem in right ∆ATO, (AT)2 +

(TO)2 = (AO)2

.

Substituting, 52 + (r – 4)

2 = r

2, and .

8

41r


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. Connect BF.

Then GH = 4, so 3)(2

1 GHDCHCDG . Since

BD = BF, triangle DBF is isosceles. So DH = HF and

CF = 4.


Draw PA so that BPC = A PC; then ∆ACP ∆A CP (asa)

and AC = A C, PA =

PA. Since PC bisects BPA in ∆BPA,

AP

PB

AC

BC

or .3

4

PA

PB

CA

BC

AB = PB – PA since A is between P and B.

,3

11

3

4

PA

PA

PA

PB

PA

AB so .3

AB

PA

Problem 14: Solution: : (C)

Drop DGBC. Let AC = x, GC = y. Note that BC = 2y, for

∆BCD is isosceles. Since ∆DCG ~ ∆ACF ~ ∆BCA, we obtain

.2

1

1

x

yx

y Thus

xy

1 and ,

2

2xy implying

x3 = 2, or 3 2x

Problem 15: Solution: (A).

Since CD trisects right angle C, BCD = 30 and DCA = 60 and CDE = 30, so that

∆DEC is a 30 – 60 – 90 triangle. Let EC = x. 3 xDE and DC = 2x.



BC

AC

DE

AE

3

4

3

4

x

x

343

12

x

Therefore13

24332

343

242

x .

Problem 16: 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 222222 BDACCDABBCAD

2222 6)84(2 BC , 1246)84(2 2222 BC 312BC .


Since CD = 60, BAO = 30. Therefore, ∆ABO is a 30 – 60 – 90

right triangle.

Since BO = 5, AO = ,35 AB = 10.

Connect CD. Since AD is the diameter, ∆ADC is a 30 – 60 – 90

right triangle. 3102 AOAD . 153102

3AC , BC = AC –

AB = 15 – 10 = 5.

Problem 18. Proof :

Method 1:

Find the middle point of CD, that is F, then connecting BF, we get DF

= CF.

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

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

=2.



Since BC = BC, BEC BFC. Therefore EC = CF. So CD = 2CF = 2CE, i.e. CD =

2CE.

Method 2:

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

AEC BEF

So 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:

Extending 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:

Extending 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 middle point of AC. Connecting BF, then CE = BF.

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



Method 6:

Extending CE to F so that CE = EF.

It is easy to prove 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 19. Proof :

Extend DA and CB to meet at F.

For right triangles DBF CAF (AAS, F =F,

DBF = CAF (right angles), AC = BD (given)).

Then AF = FB, DF = FC.

AD = DF – AF, BC = CF – FB. We already proved that AF = FB, so

AD = BC.

Problem 20: 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 midpoind of AE,

2

5

2

1 DERM inches , and hence

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.


Draw DA ⁄⁄ CA and meets the extension of BA at A.

AACD is a parallelogram with DA //AC and DA = AC.

Therefore, we know that DA B is a right angle.

222 BADBDA 2222 )()( ABDCABAABDAC



Problem 22: Solution: (C).

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

is a parallelogram. DG = 5, AF = 10, AB = 15.

Problem 23: Solution

Extend AD through D and BC through C and denote the intersection by E.

Since B = D = 90, ∆ABE and ∆CDE are right triangles. E = 30.

Therefore AE = 2AB = 8, DE = 3.

In right triangle ABE, we have: 3CD , 32CE .

In right triangle ABE, we have: 34BE , 32BC .

Hence 23:32: CDBC .


Draw whole circle and extend AB to E and DC to F to meet EF, the tangent of the circle,

at E and F, respectively. M is the tangent point.

AE = 2AB = 4, DF = 2DC = 6.

Since the trapezoid is circumscribed the circle,

AD + EF = AE + DF = 10

5)(2

1 EFADBC


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

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

xCD 3 , a = 2x, b = 90 – 2x;

3x2 + (80 – x)

2 = (90 – 2x)

2;

2

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




We try to construct two congruent triangles with correspoding sides of EG and HG.

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 triangle ABC is equilateral and B = 60, we construct an equilateral triangle BEF

by extending BD to F such that BF = BE and connecting EF.

BE = AE + AB = BD + BC (1)

BF = BD + DF (2)

Since BE = BF, BC = DF.

Therefore EBC EFD (EB = EF, B = F = 60, BC = DF).

Hence CE = DE.


Since 1 = 2, 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. 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.




We can image that the original figure is a rectangle. The

perimeter is then 232 .


If we reflect the figure about the diameter, we find that we are trying to maximum the

area of a hexagon inscribed in a circle. This hexagon will be the regular hexagon whose

sides are equal to the radius. Thus the height of the trapezoid is h = 2

35

2

3r .

50 AMC LECTURES Lecture 19 Eight More Methods To ... of AC. Find m b, the length of the median BO. Solution: Extending BO to D such that BO = OD. Connect AD and CD. Since AC and BD

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