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
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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.
Hi peter ;
DeleteAID is an isosceles triangle , Why AIO1 and DIO2 are isosceles triangles ?
How could you prove IO1 , IO2 are radius of Circle I ?
https://photos.app.goo.gl/mRovGUixhbgm9TYH6
DeleteSee attached sketch for details
CI meet arc AB at the midpoint
AI meet arc BC at the midpoint
^(DCM)= u+v
^(DIC)= ext. angle of triangle AIC= u+v
so triangle DIC is isosceles
Thanks Peter, It's clear now.
Delete