Geometry Problem

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

Click the figure below to see the complete problem 620.

## Saturday, June 11, 2011

### Online Geometry Problem 620: Three Rectangles, Diagonals, Centers, Angles

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ReplyDeletewe have KL//FC and KJ//BI

let FC cut BI at J

We have angle (IEA)= angle (DEF)=35 both angles complement to angle AED

AIBE and ECFD are cyclic quadrilaterals

so angle (DCF)= angle (DEF)= 35 ( see picture)

and angle (IBA)= angle (AEI)=35

so x=180-55-55= 70

Peter Tran

Comment(See Problem 620)

ReplyDeleteMore generally,

Let ∆PQR be the orthocentric triangle of ∆EAD

(formed by the feet P,Q,R of the perpendiculars from E,A,D upon the opposite sides AD,DE,AE respectively).

Then ∆KJL is similar to ∆PQR with

∠JKL = 180°-2∠AED,

∠KJL = 180°-2∠EAD,

∠JLK = 180°-2∠ADE.

Pravin

ReplyDeleteIt is not clear to me about statement

"Then ∆KJL is similar to ∆PQR with

∠JKL = 180°-2∠AED, "

Please explain.

Peter Tran

Let angle AED = y. Then your proof will be as follows:

ReplyDelete"we have KL//FC and KJ//BI

let FC cut BI at M (You named it as J)

We have angle (IEA)= angle (DEF)=90 - y both angles complement to angle AED

AIBE and ECFD are cyclic quadrilaterals

so angle (DCF)= angle (DEF)= 90 - y ( see picture)

and angle (IBA)= angle (AEI)=90 - y

So angle MBC=angle MCB=y

so x=180-y-y= 180-2y"

Now it is wellknown that in any triangle ABC,

if D,E,F are the feet of the perpendiculars from A,B,C on the opposite sides, and triangle DEF is formed, then its angles are given by

D=180-2A,etc

In our problem we have triangle EAD and its orthic triangle is named by me as PQR

So angle QPR = 180-2AED = angle JKL etc