Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 736 details.
Thursday, March 8, 2012
Problem 736: Triangle, Altitude, Perpendicular, Angles, Cyclic Quadrilateral
Subscribe to:
Post Comments (Atom)
http://img826.imageshack.us/img826/5130/problem736.png
ReplyDeleteQuadrilateral DBEH is concyclic
In right triangles AHB and BHC we have BD.BA=BH^2=BE.BC => quadrilateral ADEC is concyclic
Let O is the center of cycle ADEC. (See sketch)
In circle DBEH, ∠ (HBE)= ∠ (HDE)=33 => ∠ (EDC)=33-21=12
Angles at the center of circle ADEC ∠ (AOE)=2 *57=114 and ∠ (EOC)=2*12=24
So ∠ (AEB)=x=1/2*(114+24)=69
Triangles BDH and BHA are similar
ReplyDeleteSo are trangles BEH and BHC
Therefore BD.BA = BH^2 = BE.BC
So D,A,C,E are concyclic
<BDE = <ECA = 57 deg
<EDC = 90 - 57 -21 = 12 deg
<EAC = <EDC = 12 deg
Hence x = 57 deg + <EAC = 57 + 12 = 68 deg
Typo 57 + 12 = 69
ReplyDeleteThe solution is uploaded to the following link:
ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJeHNTbEVCdnk1LU0
BDHE is concyclic hence < BDE = < BHE = < BCH = 57 so ADEC is also concyclic
ReplyDeleteHence < CDE = 12 = < EAC. Therefore x =12 +57 = 69
Sumith Peiris
Moratuwa
Sri Lanka
The given angle 57 degrees is superflous. After one proves that ADEC is concyclic, then angles ADC and AEC are equal. So ADC = 90+21= 111 = AEC. Then AEH = AEC - HEC = 111 - 90 = 21. And finally x = BEA = BEH - AEH = 90 - 21 = 69 degrees.
ReplyDeleteJoaquim Maia
Brazil - Rio de Janeiro