Proposed Problem

Click the figure below to see the complete problem 340 about Triangle, Angle Bisectors, Perpendiculars, Exterior Point, Distances.

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Complete Problem 340

Level: High School, SAT Prep, College geometry

## Tuesday, August 18, 2009

### Problem 340. Triangle, Angle Bisectors, Perpendiculars, Exterior Point, Distances.

Labels:
angle bisector,
distance,
exterior point,
perpendicular,
triangle

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As in problem # 339, we assume B:(p,q), A:(0,0) & C:(b,0)and AB = c, AC = b & BC = a as usual.

ReplyDeleteSince E divides AB in the ratio b:a we've

E as(bp/(a+b), bq/(a+b))

Since D divides CB in the ratio b:c

D is ((bp+bc)/(b+c), bq/(b+c))

Slope DE =((bq/(a+b))- bq/(b+c)) /(bp/(a+b)- (bp+bc)/(b+c)) = (c-a)q/(c(b-p)+a(c+p))

DE is y = (c-a)qx/(cb+ac+pa-pc) + k and k can be determined as bqc/(cb+ac+pa-pc)since the line passes thru E by substituting co-ordinates of D in the eqn. above. Thus,DE is:

y=((c-a)qx+bqc)/(cb+ac+pa-pc)-------(1)

Let F:(x,y) be any point on DE extended satisfying the eqn. above.

Now AB is y=qx/p or qx-py=0 while AC is y=0

Slope BC = -q/(b-p)

Hence BC is y= -qx/(b-p) + k1 & it passes thru’ (b,0) so k1 = qb/(b-p). Hence BC is qx + y(b-p)-qb=0. Therefore, FG = qx-py/c since c^2 = p^2+q^2

& FH = (qx +y(b-p)-qb)/a since a^2 =(b-p)^2+q^2

We observe that FG and FH are of the same sign while FM is of the opposite sign.{If c > 0 in the equation ax + by + c = 0 of the straight line and the given point and the origin are on the same side of the given straight line, then the required length of the perpendicular is positive and if the given point and the origin are on the opposite sides of the given straight line, then the perpendicular distance will be negative}

Hence, FH -FG = (qx +y(b-p) -qb)/a - (qx-py)/c

If this is equated to FM = - y then we obtain

y = ((c-a)qx+bqc)/(cb+ac+pa-pc) which is, in fact, equation (1) above. Therefore for any point F on extended DE we’ve: FH - FG = FM .QED.

Ajit

I am interest in geometry solution not algebrica

ReplyDeletesolution. Anybody have geometry solution ?

Peter Tran

vstran@yahoo.com

Here is the geometry solution to the problem:

ReplyDelete1.Call Points M1, H1 as the projections of point E over segments AC and BC

Call Points M2, G1 as the projections of point D over segments AC and AB

Let x=EF and X=ED

2. We have EM1=EH1 ( EC is angle bisector of angle ACB)

DM2=DG1 (AD is angle bisector of angle BAC)

3.calculate FM as linear interpolation (or similar triangles)of EM1 and DM2

(EM1-FM)/(DM2-FM)= x/(X+x)

So FM= x/X*DM2 +(X+x)/X *EM1

4.Triangle FGE similar to triangle DG1E and we have

FG/DG1= x/X so FG=x/X *DG1

5.Triangle DH1E simillar to triangle DGF and we have

FH/FH1= (x+X)/X so FH= (x+X)/X *EH1

6.FH+FG= x/X *DG1 +(x+X)/X *EH1

7.Replace DG1=DM2 and EH1=EM1 we get

FH+FG=x/X *DM2+(X+x)/X *EM1= FM per step 3

So FM=FH-FG

Peter Tran

Emai : vstran@yahoo.com

See below for the sketch of pro. 340

ReplyDeletehttp://s2.postimg.org/dpyy7jsy1/pro_340.png

Peter Tran