Geometry Problem

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

Click the figure below to see the complete problem 625.

## Saturday, June 18, 2011

### Problem 625: Triangle, Parallel Lines, Congruence, Angle Bisector

Labels:
angle bisector,
Ceva's theorem,
congruence,
parallel,
similarity,
triangle

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http://img31.imageshack.us/img31/8388/problem625.png

ReplyDeleteLet CD cut AB at M and AE cut BC at L and angle bisector of angle ABC cut AC at N ( see picture)

Let AD=CE= x

we have LB/LC= x/c ( tri. LAB similar to Tri. LEC)

MB/MA=a/x ( Tri. MAD similar to tri. MBC)

NA/NC= c/a ( BN is angle bisector of angle(ABC))

Note that LB/LC . NA/NC . MC/MA= x/c .c/a .a/x= 1

So CM, AL and BN concurrent at F per Ceva's theorem and BF is angle bisector of angle ABC

Peter Tran

Denote intersection of

ReplyDeleteAF, BC by P;

BF, AC by Q;

CF, AB by R.

(AQ/QC)(CP/PB)(BR/RA) = 1 by Ceva.

Consider triangle AFC and point B.

BF intersects AC at Q; BA intersects CF at R; BC intersects AF at P

Again by Ceva, (AQ/QC)(CR/RF)(FP/PA) = 1

It follows from the above two equations that

(CP/PB)(BR/RA)= (CR/RF)(FP/PA)

Now AB // CE implies triangles CPE and BPA are similar and so

CP/PB = CE/AB

Similarly CB//AD implies triangles BRC and ARD are similar and so

BR/RA = BC/AD

Multiplying the last two equations,

(CP/PB)(BR/RA)= BC/AB (noting that CE=AD)

Thus (CR/RF)(FP/PA) = BC/AB

Now denote the areas BFC, CFA, AFB by x, y, z respectively

Clearly

CR/RF = (BRC) /(BRF)

= (ARC)/(ARF)

= [(BRC) + (ARC)] /[(BRF) + (ARF)]

=(ABC)/(AFB)

Similarly,

FP/PA = (BFC)/(ABC)

Thus (CR/RF)(FP/PA)= (BFC) /(AFB) = CQ/QA

Follows CQ/QA = CB/BA

Hence BF(Q) bisects angle ABC

Call intersection of DC and AB to be X, and intersection of AE and BC to be Y. AX/XB=DA/BC, and BY/YC=AB/CE. By Ceva, if BF cuts AC at point Z, then CZ/ZA*AX/XB*BY/YC=CZ/ZA*DA/BC*AB/CE=CZ/ZA*AB/BC=1, so CZ/ZA=BC/AB. By angle bisector theorem, BF must bisect <ABC.

ReplyDeleteLet AE, BC cut at X.

ReplyDeleteLet AD = CE = p, AF = m, FX = n, BX = t, CX = s.

From similar triangles,

p/s = m/n …(1) and

p/c = s/t ….(2)

From (1) and (2), m/n =c/t and so BF bisects <B

Sumith Peiris

Moratuwa

Sri Lanka

In the quadrilateral ABCD, AB=AD, <ABC=<ADC =90⁰. On the sides BC and CD are assigned respectively the points M and N on each sides of the triangle, such that DM⊥AN. Find that AM⊥BN

ReplyDeleteErina NJ

Hi Erina,

DeleteYour problem has been published as problem 1403 at http://www.gogeometry.com/school-college/5/p1403-quadrilateral-congruence-90-degree-angle-tutor.htm Thanks