Proposed Problem

Click the figure below to see the complete problem 409 about Quadrilateral, Diagonals, Perpendicular, Congruence.

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Complete Problem 409

Level: High School, SAT Prep, College geometry

## Saturday, December 26, 2009

### Problem 409: Quadrilateral, Diagonals, Perpendicular, Congruence

Labels:
10,
20,
30,
angle,
congruence,
perpendicular,
quadrilateral,
triangle

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there is not enough data to solve this^probleme. Values of angles ACB et ACD are irrelevant.

ReplyDeleteValues of angles ACB and CAD are relevant.

ReplyDeleteOD=a

ReplyDeleteOA=atan80

OB=atan80tan20

OC=atan80tan20tan40

tanx=OC/OD=tan20tan40tan80=tan60

x=60

.-.

Can this problem be solved without trig?

ReplyDeletebjhvash44@sbcglobal.net

Yes, this problem can be solved without trigonometry.

ReplyDeleteA synthetic Proof:

ReplyDeleteLet P be the reflection of C over BD.

Then the problem turns into Trigonometric Form of Ceva.

The angles from A in clockwise are: 10, 20, 30, 40,x, 80-x.

By a synthetic proof or as a result of trigonometric form of Ceva, 10,30,x are interchangeable; and 20,40,80-x are interchangeable.

Swap 20 and 40.

Then problem goes into 10,40,30,20,x,80-x.

In that case A=10+40=50 and B=30+20=50 and C=x+80-x=80.

Let CQ be the symmetry axis of isosceles triangle where Q is on BP.

AQP=PQC=60 and QAP=PAC=10 => P is the incenter of triangle QAC.

So PCA=PCQ=20 and PCB=x=60.

While solving this kind of problems, I always swap appropriate angles. It is a powerful tool also for synthetic solutions.

Geometric solution :http://i.imgur.com/byc89Ao.jpg?1

ReplyDeleteLet E a point ,where triangle AED is equilateral

Triangle EBD is isosceles since ∠EFA=90 and EF=FD so ∠DEC=20

Points E,B ,C are collinear since ∠BCA +∠BAC=∠EBA

EA=EC since∠ EAC =∠ECA

So Triangle DEC is isosceles so 20+40+2x=180 finally x=60

I forget say : F is point of intersection of ED and BA

DeleteAnd the point E is above AD

Excellent solution!

DeleteSlight modification.

Let DP be an altitude of Tr. ADB. Let DP and CB meet at E. Then Tr.s BEP and BDP are congruent ASA hence PE = PD and so Tr. ADE is equilateral.

Now EA = EC = ED so < EDC = 80 and < EDB = 20 so x = 60

If O is the circumcentre of triangle ABD and AO extended meets BD at E, triangle OBD is equilateral. It's easy to show that triangle DOE can be rotated clockwise by 60 degrees to coincide with triangle DBC. Hence x is 60 degrees.

ReplyDeleteReferring to your comment " It's easy to show that triangle DOE can be rotated clockwise by 60 degrees to coincide with triangle DBC. Hence x is 60 degrees. ", Can you explain further .

ReplyDeletePeter Tran

If u can draw it with the given data u can solve it at least using trigonometry

ReplyDeleteConstruct the equilateral triangle AED, like Felipe Hernandez, then <EAC=50=<ACE, so E is the circumcenter of tr. ADC and <ACD=<AED/2=30, i.e. <BDC=90-30=60.

ReplyDeleteBest regards

Problem 409

ReplyDeleteLet the point O is the center of the circle of des triangle ABD.Then triangle OBD is equilateral

And <OAB=<OBA=10,<OAD=<ODA=20,<OAC=10. I take the point K on the side AC such that

<DKC=40,so <KDO=10=<KAO , then AOKD is cyclic,then <KOD=<KAD.Therefore KO=KD (OB=BD) so BK is perpendicular bisector to OD.Then <DKB=<BKO=90+10=100,so

<CKB=100-40=60. But <DKC=40=<DBC so DKBC is cyclic.Therefore < CDB=<CKB=60.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Problem 409

ReplyDeleteI take the point A symmetry of A on the AB, then triangle AED is equilateral (AE=ED=AD)and <EBA=<ABD=70, but <EBA+<ABD+<DBC=70+70+40=180.So E,B and C are collinear.

Is <EAC=<ECA=50 then AE=EC .So the point E is the center of the circumcircle of the triangle CAD.Then <ACD=(<AED)/2=60/2=30. Therefore <BDC=90-30=60.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE