See complete Problem 47

Angles, Triangle, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 47

Labels:
angle,
congruence,
triangle

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good problem

ReplyDeletewe have to draw a perpendicular from from B to AC and E is the foot of the perpendicular .prove triangle BED is a rightangled isoscleles triangle

hence the result

Draw BE,CF perpendicular to AC & AB extended respectively. Let AD=DC=p. From Tr. ACF we've CF = AC/2 = p while angle FBC =30+15=45; hence BF=p and from Tr. ACF, AF=2p*V3/2 =V3*p. Thus, AB=V3p-p. Now angle ABE=60 and so BE=(V3p-p)/2 and AE=(V3p-p)V3/2 which gives DE = p-(V3p-p)V3/2 =(V3p-p)/2. In other words BE = DE or angle DBE=45 deg. Finally, angle ABE + angle DBE + x + 45 = 180 or 60 + 45 + 45 + x = 180 which yields x = 30 deg.

ReplyDeleteAjit: ajitathle@gmail.com

Prolong AB, then draw a perpendicular from C to AB, being F de foot of this perpendicular.

ReplyDeleteThen triangle DFC is euqilateral and BDF is isÃ³sceles of base BD.

Completing angles in the diferents triangles we'll have in BDF that 30 + 2(45+x)=180 <=>2x=60 <=> x=30

Greetings :D!

Drop a perpendicular CE to AB extended. Then Tr. BCE is right isocoles and BC^2 = 2BE^2 = CD.CA

ReplyDeleteHence BC is tangential to Tr. ABD and so x = 30

Sumith Peiris

Moratuwa

Sri Lanka

Or let E be the foot of the perpendicular from C to AB and let F be on AE extended such that BE = EF

ReplyDeleteThen AD = DC = DE = BE = CE = EF

So < BDF = 90 = < BCF. Hence BDCF is cyclic and < BFC = 45 = < ADB

Hence x = 30

Take E reflection of C about AB, tr ACE is equilateral, <CBE=<CDE=90, BDCE is cyclic, hence <CBD=<CED=30, done.

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