Geometry Problem

Problem submitted by Charles T.

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

Click the figure below to see the complete problem 832.

## Friday, December 14, 2012

### Problem 832: Quadrilateral, Isosceles triangle, 70-70-40 degrees, Congruence, Auxiliary lines, Equilateral triangle

Labels:
angle,
au,
congruence,
equilateral,
isosceles,
quadrilateral

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http://img51.imageshack.us/img51/5048/problem832.png

ReplyDeleteFrom D draw DE//BC and DE=BC=AD

BCDE is a parallelogram => BE //DC => BE is angle bisector of angle ABD

So ED=AD=AE => ADE is a quadrilateral triangle and ∠ ADE=60

So ∠BDC=∠BDE=70-60=10

Anonymous

DeleteWithin angle ABD build angle equilateral APD , BP perpendicular bisector for the segment AD .

BP parallel with DC , DP=BC than BPDC trapezoid or parallelogram,

BPD=150grade when ABC <90 we have parallelogram, and angle DPC is 10 grade .

When 90<ABC<180 it becomes that DPC=130 grade

Erina New Jersey

Complete the rectangle ADCE. Join BE.

ReplyDeleteTriangle BEC is equilateral.

Since triangles BEA,BCD are congruent,

angles ABE,DBC are equal each equal to

(60 - 40)/2 = 10 degrees.

Construim paralelogramul ADMB si vom triunghiul echilateral BMC deoarece triunghiurile ABC=BDM si BCD=MCD=>m(<CBD)=m(<MBD)-m(<CBD)=70-60=10

ReplyDeleteOn the angle bisector of B mark E such that BE = CD. But BE is also parallel to CD hence BECD is a parelleligram and DE = and is oarellel to BC = AE SAS

ReplyDeleteHence Tr. AED is equilateral and x = < EDB = 10

Sumith Peiris

Moratuwa

Sri Lanka