See complete Problem 108

Triangle, Angles, Median, Congruence, Cyclic Quadrilateral. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 108

Labels:
angle,
congruence,
cyclic quadrilateral,
median,
triangle

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Think about a cyclic quadrilateral.

ReplyDeleteGeometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

Draw a perpendicular through D, wich intersects BC in E. Then triangle DEC is congruent to triangle DEA. That implies that DEA=90-alpha, wich shows that ABED is cyclic, so EBD=EAD. this is alpha=x.

ReplyDeletesrry for my english :$

Maybe you meant:

DeleteDraw a perpendicular through D, which intersects BC in E.

Then the triangle DEB is congruent to the triangle DEA.

That implies that DEA=90-alpha

which shows that ACED is concyclic,

so EAD=alpha=ECD=x.

Therefore is alpha=x.

This problem x has two solutions

ReplyDeletex=Alpha

x=90-Alpha

http://geometri-problemleri.blogspot.com/2009/06/problem-29-ve-cozumu.html

ReplyDeleteLet the perpendicular bisector of AC meet AB extended at E. Then < EAC = 90-x and so < AED = x = < DEC = < DBC

ReplyDeleteHence BDCE is cyclic and ABC is a right triangle with D the centre and so x = @

Sumith Peiris

Moratuwa

Sri Lanka

Draw a circle passing through A, B and C

ReplyDeleteDefine E as intersection of CD and this circle

ang(CDA)= a+x => ang(DAC)= 90-x

ang(ECB)= x => ang(BAE)= x (intercepts the same arc BE)

Therefore ang(CAE)= 90 and the center O of the circle is on CE

Let’s suppose that D is not the center of the circle but O

We already know that O is on CE

On a circle, if we take two points, A and B, then the center of the circle is on the perpendicular bisector of AB

Therefore, the center of our circle is on the perpendicular bisector of AB

The intersection of the perpendicular bisector of AB and CE is D

Therefore D is the center of the circle and DC=DB, ΔCDB is isosceles

Therefore x=a

[For easy typing, I use "a" instead of alpha]

ReplyDelete<BAC=90-x

Consider triangle BDC

sinx/CD=sina/BD

CD/BD=sinx/sina------(1)

Consider triangle BDA

sin(90-a)/AD=sin(90-x)/BD

AD/BD=sin(90-a)/sin(90-x)-------(2)

Since AD=DC, by equating (1) & (2)

sinx/sina=sin(90-a)/sin(90-x)

sinx/sina=cosa/cosx

sinacosa=sinxcosx

sin2a=sin2x

2a=2x

a=x