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.

## Thursday, April 23, 2020

### Geometry Problem 1472: Cyclic Quadrilateral, Perpendicular Diagonals, Rectangle

Labels:
concyclic,
cyclic quadrilateral,
diagonal,
GeoGebra,
ipad,
ipadpro,
perpendicular,
rectangle

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∆APE1, ∆PBE1 isosceles => E1 midpoint, ∆PCE2, ∆PBE2 isosceles => E2 midpoint

ReplyDelete=> E1E2 midlle line of ∆ABC , at the same three rest sides

https://photos.app.goo.gl/9X2XxujZ82wfkgDy9

https://photos.app.goo.gl/UCbnTTv6vi2aEGSK6

ReplyDelete< E1BP = < PCD = < DPH3 = < BPE1 so BE1 = PE1 and so E1 is the midpoint of AB

ReplyDeleteSimilarly E2, E3 and E4

Hence E1E2 // AC // E3E4 & similarly E1E4//BD//E2E3

Therefore E1E2E3E4 is a parallelogram & since AC//BD, it follows that E1E2E3E4 is a rectangle

Sumith Peiris

Moratuwa

Sri Lanka

AC perpendicular to BD and not AC//BD in the last line of my proof above

Delete