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.

## Wednesday, February 26, 2020

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

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Let AOH cut circle O at X

ReplyDelete< 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

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

Delete-Pulliwaya

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

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

Delete-Pulliwaya

No worries Pulliwaya

Deletehttps://photos.app.goo.gl/Uhb23htgvw5KE5q69

ReplyDeleteConnect 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

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

DeleteYes, You are right .See link below for detail .

Deletehttps://photos.app.goo.gl/47JzfogMBJ6stvkUA