Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1230.

## Tuesday, July 5, 2016

### Geometry Problem 1230: Quadrilateral, 60 Degrees, Congruence, Metric Relations

Labels:
60 degrees,
congruence,
metric relations,
quadrilateral

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https://goo.gl/photos/TSKinFA2FzwTcMNU6

ReplyDeleteLet AB meet DC at K

AKD is an equilateral triangle

Draw BF // AD and let KE meet AD at X

We have KB=2.a and CD=2.a

We have BD^2= a^2+9a^2- 2.3a.ạ cos(60)= 7.a^2 => BD= a√7

Apply Ceva’s theorem in triangle AKD we have

BA/BK x CK/CD x XD/XA = 1 => XD/XA= 4

We also have ED/EB= XD/XA + CD/CK= 4+ 2= 6

So ED/6=EB/1=BD/7 => EB= BD/7= a/√7

https://goo.gl/photos/baBUwbeKwDLUEN1V9

Delete(1) KBF is equilateral and AC=AF=BF= a√7

Apply Van Aubel theorem for point E in triangle KAD

AE/EC=BA/BK+XA/XD= ¾

So AE/3=EC/4=AC/7=a/√7

And AE= 3.a/√7 => BE/AE=1/3

(2) .. In equilateral triangle KBF, the median BC perpendicular to KF

(3)…BC= KB. √3/2= a√3

(4).. S( ABCD)=S(KAD)- S(KBC)

S(KAD)= AD^2. √3/4

S(KBC)=SB^2. √3/8

Replace AD= 3.a and SB=2.a we get S(ABCD)=7/4. a^2. √3

There is no explanation why cd=2a

DeleteNote that

Delete- triangle AKD is equilateral

- FD= a

- triangle FAC is isosceles ( FA=BD=AC)

from above information we can calculate CD= 2a

Oops, you used trigonometry. Ah well.

DeleteNo need of Trigonometry to prove CD = 2a.

DeleteSince AC = AF, triangles AKC and ABD are congruent ASA.

Hence KC = AB = a.

So CD = 2a

Make F such that AFD is equilateral (true as angle BAD and ADC are 60 degrees).

ReplyDeleteMake H on FD such that AH = BD, BH is parallel to AD. (Feasible as triangle AFD is equilateral which are symmetrical about their altitudes).

Make G on FD such that AG is perpendicular to FD. As AFD and ACH show properties of isosceles triangles, G is bisector of CH and FD.

Hence, CD = 2a.

Mass points: Let E be the center of mass if fixed masses are placed on A, F, D (lines are frames which are strong and massless).

Let D have mass 1. As E is center of mass, B (which is concurrent with E and D) must be center of mass of AF. Similarly, C is center of mass of FD.

As CD = 2a, FC must be a, so the mass on F must be 2.

As AB = a, BF must be 2a and therefore the mass on A is 4.

So the total mass of AF is 6. This means that mass ratio of B:D is 6:1, so the length ratio of BE:ED = 6:1

This indicates that BE has 1/7 of length of BD.

BD = AC, AC^2 = AG^2 + CG^2 = AD^2 - GD^2 + CG^2 = 7*a^2

Hence BD = sqrt(7) a

That means that BE = a/sqrt(7)

I said in an earlier post that F has mass 3, D is mass 1 and A is mass 3. This indicates that FD has total mass of 3.

ReplyDeleteSo, mass ratio A:C = 4:3, which means that AE:EC = 3:4

Hence, AE is 3/7 of AC.

I also found in that other post that AC is sqrt(7) a. This means at AE has length of 3*sqrt(7)*a, which is thrice of BE.

whoops, at the end, I meant to say "AE has length of 3*a/sqrt(7)" which is really thrice of BE.

DeleteExtend AB to meet DC at P and form the equlateral triangle APD

ReplyDeleteSince given AD = 3a and AB = a

we have BP = 2a, PC = a, CD = 2a and since m(BPC) = 60

=> BPC is a 30-60-90 triangle

Hence BC = √3a

Since m(BCP) = 90 => m(BCD) = 90 (Hence BC _|_ CD)

Applying pythogrous to triangle BCD

=> BD^2 = BC^2+CD^2

=> BD^2 = 3a^2+4a^2

=> BD = √7a = AC

Observe that the triangles PAC and ADB are congruent per SSA (since PA = AD, AC=BD and m(APC) = m(BAD) = 60)

Let m(ADB)=m(PAC)= α

=> m(PCA) = 120-α => m(BCA) = 30-α

and m(BDC) = 60-α => m(DBC) = 30+α (Corrected this from m(DBA) to m(DBC))

∴ in the triangle BEC, m(BEA) = 120

=> m(BEA) = 60

So triangle EAB is similar to ADB

∴ AB/BE = BD/AB

=> BE = a^2/√7a

Hence BE = a/√7

Similarly AE/BE = AD/AB = a/3a = 1/3

Hence AE = 3BE

Finally, Area of Equilateral triangle APD = √3(9a^2)/4

Area of right triangle PBC = √3(a^2)/2

∴ Area ABCD = Area APD - Area PBC = √3(a^2)[9/4-1/2] = 7√3(a^2)/4