## Monday, August 27, 2018

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

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

Details: Click on the figure below.

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

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

2. nice question [ zero sir from Nepal ]

3. Let the length of altitude from E to AD = H and E to BC = h

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

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

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

5. Solution without using altitudes.

Draw 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