Geometry Problem 1491. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

## Tuesday, December 8, 2020

### Geometry Problem 1491: Cyclic Quadrilateral, Diagonal, Incircle, Angle, Measurement

Labels:
angle,
circle,
cyclic quadrilateral,
diagonal,
geometry problem,
incircle,
measurement

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I think there is a typo and O1 is the Incentre of Tr. ABD not ABC

ReplyDeleteLet <ABD = <ACD = 2p.

Its not difficult to infer (considering Tr. ABD) that <AO1D = 90+p = <AO2D (considering Tr. ACD)

Hence AO1O2D is concylic

Again considering Tr. AO1D, <ADO1 = 180 - (90+p) - 40 = 50 - p = <AO2O1

So <O1O2D = (50-p) + (90+p) = 140

Sumith Peiris

Moratuwa

Sri Lanka

If we are given the lengths of the 4 sides of ABCD and the length of either diagonal we can easily calculate the length of O1O2 (irrespective of the angle A)

DeleteJoin A to O1. Quadrilateral AO1O2D cyclic

ReplyDelete=> Ang O1O2D=140

See sketch for details

https://photos.app.goo.gl/BezctJeJzUHUpM9y8

ReplyDeletehttps://photos.app.goo.gl/ZYD4fjg9k2jNDHAF9

ReplyDeleteLet BO1 and CO2 meet at I

I will be the midpoint of arc(AD)

We have isosceles triangles AID, AIO1, and DIO2

In circle ABCD, ∡ (BID)= ∡ (BAD)= ∡ (DIO1)= 80 deg.

In circle I, arc(O1AD)= 360-∡ (DIO1)=280 deg.

So ∡ (O1O2D)= ½ x 280=140 deg.