## Wednesday, February 26, 2020

### Dynamic Geometry 1454: Intersecting Circles, Perpendicular Lines, Cyclic Quadrilateral

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Let AOH cut circle O at X

< XBD and < DBA are complementary angles
But < XBD = < HAE and < DBA = < HEA

Hence < HAE and < HEA are complementary and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

1. Sumith - Could you please care to explain how XBD and DBA are complimentary

-Pulliwaya

2. They add upto < XBA which is 90 degrees since XA is a diameter of Circle O

3. Thank you for the response sir. My bad I imagined X on big circle and hence the confusion.

-Pulliwaya

4. No worries Pulliwaya

2. https://photos.app.goo.gl/Uhb23htgvw5KE5q69

Connect CD, OQ, OD and QD
Let F is the midpoint of CD
In circles O and Q we have ∠ (CAD)= ∠ (QOD) and ∠ (ACD)= ∠ (OQD)
So Triangle CDA similar to QDO ( case AA)
So ∠ (QDF)= ∠ (ODA)= ∠ (OAD)
On circle Q we have ∠ (CDE)= ∠ (FQD)
So triangle FQD similar to HEA (case AA)
So ∠ (AHE)= ∠ (DFQ) = 90 degrees

1. shouldn't line 6 read "on circle Q we have <(CED)=<(FQD)?

2. Yes, You are right .See link below for detail .
https://photos.app.goo.gl/47JzfogMBJ6stvkUA