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

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

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## Tuesday, September 26, 2017

### Geometry Problem 1347: Triangle, Angle Bisector, Congruence, Circumcircle, Tangent, Equilateral Triangle

Labels:
angle bisector,
circle,
circumcircle,
congruence,
equilateral,
geometry problem,
tangent,
triangle

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∠BDC = ∠BCE = ∠BEC, so BDCE is concyclic.

ReplyDelete=> ∠ADE = ∠ACB

=> ∠ADE = ∠EDC = ∠BDC = 60.

Please, explain your first step. Thanks

DeleteI thought it is obvious so I skip the explanation.

Delete* ∠BDC = ∠BCE is because the triangle BDC & triangle BCA are similar under the power of circle at B (with C is the tangent point),

** ∠BCE = ∠BEC is because BE = CE so that triangle BCE is isos.

Let X be the foot of the perpendicular from E to AB and Y that of E to DC.

ReplyDeleteTriangles EXD and EYD are congruent ASA.

So EX = EY and hence triangles EXB and EYC are congruent right angle, hypotenuse, side.

It follows that < DBE = < DCE and hence BCED is concyclic and so < C = < BCE = < EBC = < BDC = < BEC

Therefore Tr. BCE is equilateral

Sumith Peiris

Moratuwa

Sri Lanka

To Sumith

DeleteI have 2 comments :

1. Your proof never use the given data " BC tangent to the circumcircle of triangle ADC at C "

2. referring to line 4 of your solution "

It follows that < DBE = < DCE and hence BCED is concyclic and so < C = < BCE = < EBC = < BDC = < BEC "

Please explain how do you get < EBC=<BDC

Peter Tran

Peter

DeleteYour 2 comments are interconnected.

Because of the tangency,

<A = DAC = DCB and hence Tr.s ABC & CBD are similar and so

< BDC = C.

But < EBC = < ECB = C

Hence < EBC = < BDC

Hope this clarifies

Sumith

Once you prove BCED is concyclic, without additional construction we can prove BEC as equilateral triangle by using alternate-segment theorem.

DeleteSuppose say the m(BCA) = x => m(ADC) = 180-x => m(CDB) = x

Since BCED is concyclic, m(CEB) = m(CDB) = x and hence BCE is equilateral triangle.

Repeating my response to Peter's comments

Delete1- I do, in fact,use the data on the tangency which gives < A = < BCD

2 - As a result of the 1- above Tr.s of ABC and CBD are similar. Hence < BDC = <BCE = < EBC.

Trust this clarifies

Sumith

Thanks for the clarification

DeletePeter Tran

https://photos.app.goo.gl/UZHIJKvyxwGfqzUG3

ReplyDeleteLet ∠ (ADE)= ∠ (EDC)= u

Let X and Y are the projection of E over AD and DC

Triangle BEX congruent to CEY => ∠ (XBE)= ∠ (ECY) => quadrilateral BDEC is cyclic

Let DE cut circumcircle of triangle ADC at F .

F will be the midpoint of arc AFC and ∠ (FAC)= ∠ (FCA)= u

Since BC tangent to circle ADC => ∠ (AFC)= ∠ (ACB)= u

So AFC is a equilateral triangle and u= 60 degrees

So EBC is also a equilateral triangle

Let's call ang(BAC)=y, ang(ECB)=ang(BEC)=u, and ang(ADE)=ang(EDB)=x.

ReplyDelete1. ang(CBD)=y because of BC tangent.

2. Notice in the intersection of CE and DB that 180°-u-y=180°+u-2x-y. Then we have u=x.

3. DCBE is a cyclic quadrilateral because diagonal angles in C and D are equal.

4. Because of (3), ang(CED)=ang(CBD)=y and ang(DCE)=ang(DBE)=x-y.

5. From (4) we have that ang(CBE)=ang(CBD)+ang(DBE)=x=u. Which proves that CEB is equilateral. Q.E.D.