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

Labels:
altitude,
angle,
cyclic quadrilateral,
perpendicular,
triangle

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