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.

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## Sunday, March 12, 2017

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

Labels:
angle bisector,
chord,
circumcircle,
secant,
triangle

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Draw circle O1 throw A,W,Y,B. Extend AX to P, WX to Q (P,Q on O1)

ReplyDeleteAng PWY = α, need to prove PWQ = β

ZX//WP => <PWQ = β

Problem 1322

ReplyDeletelet'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

Draw WV // ZX (with V on BC).

ReplyDeleteSo <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.

Easy to observe A,B,Y and W are concyclic.

ReplyDeleteDenote <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

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.

ReplyDelete