Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Tuesday, August 2, 2016
Geometry Problem 1243: Quadrilateral, Four Exterior Angle Bisectors, Semi-sum, Angles. Mobile Apps
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ReplyDeleteLet BC and AD when extended meet at P. We have shown earlier (see my proof
of Problem 1242) that A1,C1,P are collinear points and that this line is the bisector of <P = p say.
Consider Tr. PCC1 where <OC1B1 = p/2 + (90 - â/2)..…(1)
Consider Tr. PAA1 where <OA1D1= (90 – á/2) – p/2….(2)
A1B1C1D1 being cyclic (Problem 1241), from (2), <OB1C1 = (90 – á/2) –
p/2….(3)
Now consider Tr. OB1C1 where è + {(90 – á/2) – p/2} + { p/2 + (90 - â/2)} =
180 from (1) and (3)
Simplifying, è = (á+â)/2
(Sorry about the different symbols for alpha, beta and theta)
Sumith Peiris
Moratuwa
Sri Lanka
https://goo.gl/photos/ZXhcHx3B542E7nND7
ReplyDeleteLet BC meet AD at P and AB meet CD at Q
In triangle CDP, CC1 and DC1 are angle bisectors of angles C and D
So PC1 is an angle bisector of angle P
In triangle ABP, BA1 and AA1 are external angle bisectors of angles B and A
So PA1 is an angle bisector of angle P => P, C1 and A1 are collinear
Similarly Q, B and D are collinear.
In triangles OCP and OCQ we have β = angle (PCQ)= θ+x+y
In triangles OAP and OAQ we have α= angle ( BAD)= θ-x-y
Add above expressions side by side we have θ= ½( α+ β)