Geometry Problem

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

Click the figure below to see the problem 722 details.

## Friday, January 27, 2012

### Problem 722: Squares, Circumscribed Circles, Collinearity

Labels:
circumscribed,
collinear,
intersection,
square

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http://img832.imageshack.us/img832/7080/problem722.png

ReplyDeletein circle O’ angle(HDF)= angle (HEF)---- (both face the same arc)

in triangle OO’D angle (HDF)= angle (O’OD)= ½ angle (DOH) = angle (HAD)

so angle (AED)+ angle (HEF)= angle (AED)+ angle (HAD)=90

so angle (AEH)= angle (DEF)+90=180 >> A,E,H collinear

Peter, the following statement is true only if A,E & H are collinear

ReplyDelete"in triangle OO’D angle (HDF)= angle (O’OD)= ½ angle (DOH) = angle (HAD)"

How can we assume this to be true? Kindly explain.

The circles (O) and (O’) are orthogonal

ReplyDeletesince ∠ODO' is a right angle,

being the sum of ∠ODC and ∠O'DE,

each of which is 45°.

BHF is a straight line, since diameters BOD and DO'F

each subtends a right angle at H.

Thus in the right ΔBDF, we have DH ⊥ BF.

So ΔDHF and ΔBHD are similar.

Let ∠EAD be x.

Let a, b be the side lengths of

the squares ABCD and DEFG resp.

tan ∠HEF = tan ∠HDF = tan ∠DBH

= tan ∠DBF = DF / DB

= b √2 / a √2 = b/a

= ED/AD = b/a = tan x

Follows ∠HEF = x = ∠EAD and

A, E, H are collinear.

http://img31.imageshack.us/img31/8017/problem7221.png

ReplyDeleteThanks Ajit for your comment.

Below is my clarification and correction of my previous solution ( see picture for detail)

in circle O’ angle (HDF)= angle (MDO’)=angle (HEF)---- (both angles face the same arc)

Note that triangles OO’D, AED and DEF are right triangles and OO’ is the perpendicular bisector of DH

Triangles AOD and DO’E are isosceles right triangles and DE/DO’=AD/DO= sqrt(2)

And triangle ODO’ similar to triangle ADE ---- (case SAS)

So angle (O’OD)= angle (EAD)= angle (MDO’)= angle (HEF) and angle (AED)= angle (MO’D)

Since angle (MDO’) supplement to angle (MO’D) so angle (HEF) supplement to angle (AED)

so angle (AEH)= angle (DEF)+90=180 >> A,E,H collinear

A much simpler, albeit less elegant proof goes like this:

ReplyDeleteIn triangle DHE, angle CHE is 90° becaue C,HG are collinear (previous question) and angle AHG is 90°.

Drawing triangle ACH yields that angle DHA=90° (AC diameter, H point on circumf)

So we have CHA =90° and CHE=90° which implies the E lies on AH

ABCH is cyclic hence <EHC=90.

ReplyDelete<BAE=90-<EAD implying that<ECH=180-90+<EAD-90=<EAD

Therefore,

<ECH=<EAD

<EDA=<EHC=90

This implies that <CEH=<AED.

A,E,H are collinear

< EHD = < EFD = 45

ReplyDeleteIn cyclic quadrilateral ACHD < ACD = 45 = < AHD = 45. But < EHD = 45. So AEH must be collinear

Sumith Peiris

Moratuwa

Sri Lanka

From this it is also easy to see that CHG are collinear

ReplyDelete<HEF=<EAD (corr. <s EF//AG)

ReplyDelete<EAD+<AED=90

<HEF+<AED=90

<HEF+<AED+<DEF=180

AEH is a st. line