See complete Problem 89

Triangle area, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Monday, May 19, 2008

###
Geometry Problem 89

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 89

Triangle area, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

Let X be the mid-pt of BC,

ReplyDeleteXM//BA and XN//CD,

S_1=[EBX]=S_3/2=[EXC]=S2

let area of ABC be S

ReplyDeletearea of CEN = EC/ED*[END]

[END]= 1/2[BED]

[BED]=BE/AE*S

Thus, [CEN]=1/2*S*(EC/ED)*BE/AE

[BEM]=BE/AE*[AEM]

[AEM]=1/2*[AEC]

[AEC]=EC/ED*S

Thus,[BEM]=1/2*S*(EC/ED)*(BE/AE)=[CEN]

[BEC]=(BE/AE)*(EC/CD)*S

Therefore, S1=S2=1/2S3

Here is an alternative solution to problem 89.

ReplyDeleteLet CP and MQ be altitudes of triangles BEC and EBM respectively. In triangle APC, M and Q are midpoints of the sides, so MQ = CP/2,

hence S1 = S3/2.

Let BR and NS be altitudes of triangles BEC and ECN respectively. In triangle DBR, N and S are midpoints of the sides, so NS = BR/2,

hence S1 = S2/2.

Thus S1 = S2 = S3/2.

Area triangle MBE is (MBE)=S1=(AEM)-(ABM)=(AEC)/2-(ABC)/2=(BEC)/2=S3/2.But S2=(NEC)=

ReplyDelete(NED)-(NCD)=(BED)/2-(BCD)/2=(BEC)/2=S3/2.