Geometry Problem. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Monday, August 27, 2018

### Geometry Problem 1381: Parallelogram, Exterior Point, Triangle, Area

Labels:
area,
exterior point,
geometry problem,
parallelogram,
triangle

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h the altitude of ABCD from B, h1 the altitude of BEC from E

ReplyDeleteS4-S2=(AD(h+h1))/2 - (BC h1)/2 = (AD h)/2 = S/2

H1, H2 the altitudes of ABE, DEC from E to AB, DC

S1 + S3 = (AB H1)/2 + (DC H2)/2 = (AD H)/2 = S/2

The fourth row need to be =(AB H)/2 (not (AD H)/2

Deletenice question [ zero sir from Nepal ]

ReplyDeleteLet the length of altitude from E to AD = H and E to BC = h

ReplyDeleteS4-S2=AD*(H-h)/2=S/2 -----(1)

Let AE intersect BC at F and ED intersect BC at G

Let Area of BEF=S(BEF)=a and similarly, S(EFG)=b,S(EGC)=c,S(BFA)=d,S(AFGD)=e,S(DGC)=f

From (1)

=> (e+b)-(a+b+c)=S/2

=> e = S/2+(a+c) ---------(2)

Also d+e+f=S

From (2)=> (d+a)+(f+c)=S/2

=>S(ABE)+S(DEC)=S/2 ------------(3)

From (1) & (3)

S1+S2=S4-S2=S/2

https://photos.app.goo.gl/fhhyf1JeGS28qM6J8

ReplyDeleteDraw altitudes BF, EK, EH , EL and EM ( see sketch)

Note that BLMF is a rectangle

And S= CD x BF= AD x HK

S1+S3=1/2( AB x EL+ CD x EM)= 1/2AB(EL+EM)= S/2

S4-S2= ½(AD x EH-BC x EK)= 1/2AD(EH-EK)= S/2

So S1+S3= S4-S2= S/2

Solution without using altitudes.

ReplyDeleteDraw parallel thro E to BC thereby completing parallelogram AXYD. Complete parallelogram BXEF, F on BC

S(XYDA) = 2S4 and S(XYCB) = 2S2 and subtracting we have

S4 - S2 = S/2 .......(1)

S4 = S1 + S3 + S2 .....(2) since Tr. BEC is congruent with Tr. AFD

From (1) and (2) the result follows

Sumith Peiris

Moratuwa

Sri Lanka