Geometry Problem

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

Click the figure below to see the complete problem 835.

## Sunday, December 23, 2012

### Problem 835: Isosceles Triangle, Double Angle, Triple Angle, Auxiliary Lines

Labels:
angle,
auxiliary line,
double angle,
isosceles,
triangle

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Result: 150

ReplyDeleteSolution: Let BC=AC=1 a:alfa, b:beta

From sinus theorem in BDC

DC=sina/sin(a+b)

Sinus theorem in ADC using DC we get

sina*sin(2a+3b)=sin2a*sin(a+b) since sin2a=2*sina*cosa we get

2*sin(a+b)*cosa=sin(2a+b)+sinb=sin(2a+3b)

sinb=sin(2a+3b)-sin(2a+b)=2*sinb*cos(2a+2b)=sinb

eliminating sinb(b can't be zero) we get

cos(2a+2b)=1/2 which means 2a+2b=60, a+b=30

So x=180-30=150 degrees.

And using elementary geometry?

Deletehttp://img38.imageshack.us/img38/3597/problem835.png

ReplyDeleteMake drawing per attached sketch

1. Draw altitude CF and bisector AF of angle DAC

Make angle ACE= beta

2. Due to symmetric properties , CDE and AEDB are isosceles

And BE, AD and CF are concurrent at point G

3. In triangle AGC , GE is an angle bisector => angle BGF=angle AGF=angle AGE=angle CGE=60

In triangle ABC we have 60+ 4alpha+4.beta=180 => alpha+ beta=30

In triangle BDC x=180-alpha-beta= 150

not clear at all. Your proposition is to

Delete1) Drop an altitude from C to the base, intercepting the base at F ( I think) that will bisect the vertex angle at C into 2 equal angles of 2beta each

2) The second construction I am not sure I understand. but it seems that you are proposing and angle bisector from A intersecting BC at E or intersecting DC At E.

the rest pointing to concurrency at G etc. I have no clue at all.

do you have a sketch of your construction?

Minor typo correction.

ReplyDeleteLine 3 of my solution should read as "1. Draw altitude CF of triangle ABC and bisector AE of angle DAC "

Peter Tran

Fie P punctul de pe segmentul AD astfel incat m(: PC este mediatoarea segmetului AB si bisectoarea unghiului =>m(In ∆BPC

ReplyDeletem(6α+6β=180 =>α+β=30 de unde in ∆BDC x=180-(α+β)=150

Let CX be the bisector of < ACB, X being on AD. Now thro' congruence or symmetry < AXC = < BXC. But D is the in centre of Tr. ACX, so XD bisects < BXC. It thus follows that < BXD = 60.

ReplyDeleteSo 2 alpha + 2 beta + 120 = 180 in Tr. ADC, implying that alpha + beta = 30.

So from Tr. BDC x = 150

Sumith Peiris

Moratuwa

Sri Lanka