## Sunday, March 12, 2017

### Geometry Problem 1322: Triangle, Angle Bisector, Circumcircle, Chord, Secant, Sum of two Angles

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Details: Click on the figure below..

1. Draw circle O1 throw A,W,Y,B. Extend AX to P, WX to Q (P,Q on O1)
Ang PWY = α, need to prove PWQ = β
ZX//WP => <PWQ = β

2. Problem 1322
let's say that <BAX=δ=<XZB(B,X,Z,A=concyclic) and <ΒWX=γ,but <XBZ=α=<YBW=<XAZ=<ZAW=<YAW.So the point A,B,Y and W are concyclic.Τhen
<BWY=<BAY or γ+θ=δ+α (δ=β+γ in triangle ZXW) or θ=α+β.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

3. Draw WV // ZX (with V on BC).
So <XWV = β (alternate angles).
Enough to show <YWV = α.
A,B,Y,W are concyclic (Note <WAY = <ZAX = <ZBX = α).
So <YVW = <YXZ = <BAY = <BWY.
Hence Triangles YVW, YWB are /// and <YWV = <WBY = α.

4. Easy to observe A,B,Y and W are concyclic.
Denote <ABW = <AXZ = <AYW = γ
Let AY and XW intersect at V.
We evaluate <VYW in two ways:
As an exterior angle of triangle VYW (i.e.) θ+γ
As an exterior angle of triangle VXA (i.e.) (α+β)+γ
Follows θ=α+β

5. Easy to see, ABXZ, ABYW are cyclic, hence <AZX=<AWY=180-<ABC ( 1 ). Since <AZX=<AWX+<ZXW+<ZAW, we get the required <XWY=<ZXW+<ZAW.