## Thursday, April 16, 2015

### Geometry Problem 1110: Right Triangle, External Squares, Catheti, Cathetus, Angle Bisector, 45 Degree, Internal Square

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

1. AQ/QC = AB/BC = AB/CF = AM/MF
Thus QM//CF//AB.

Similarly QN//AE//BC.

Since BM=BN, so BMQN is square.

2. Construim perpendiculara in M pe BC si notam cu P intersectia ei cu AC.Aratam ca P=Q adica MP=BM=NB=x,trENA~trCNB=>c/a=(c-x)/x=>cx=a(c-x);PQ||AB=>trCMP~trCBA=> (a-x)/a=PM/c <=>
1-x/a=PM/c <=>c-cx/a=PM=>c-a(c-x)/a=PM=>PM=x,unde a=BC,c=AB,x=BN,=>BMPN-patrat(din problema 1109 avemBN=BM) si in final P=Q,BQ este bisectoare

3. We call BN=Z ; BM=Y NQ=X, AB=c; BC=a
First step: We have: Z/c=a/(a+c) hence Z= ac/(a+c)
Y/a=c/(a+c) hence Y=ac/(a+c) so Y=Z
Second step: ( a-Y)/X=X/(c-Z) ; X2= (a-Y)(c-Z) =(a-ac/(a+c))(c-ac/(a+c))
And following X2= a2.c2/(a+c)2 and X=ac/(a+c) =Y=Z and consequently MQNB is a square

4. Although similar to Jacob Ja's, CQ/AC=BC/(BC+AB)=BC/CD=CN/CE, hence the triangles ADE and QBN are homothetic, therefore similar; likewise CFG and QMB, therefore we are done.

Best regards