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

Level: Mathematics Education, High School, Honors Geometry, College.

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## Monday, August 1, 2016

### Geometry Problem 1241: Quadrilateral, Four Exterior Angle Bisectors, Cyclic Quadrilateral, Circle

Labels:
bisector,
circle,
cyclic quadrilateral,
exterior angle,
quadrilateral

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< A1AB = 90 - A/2 and <A1BA = 90 - B/2

ReplyDeleteHence <A1 = A/2+B/2….(1)

Similarly we can show that

<C = C/2+D/2….(2)

So <A1+<C1 = (A/2+B/2)+(C/2+D/2) = (A+B+C+D)/2 = 360/2 =180

Hence A1B1C1D1 is a cyclic quadrilateral.

Sumith Peiris

Moratuwa

Sri Lanka

It can be proved that the diagonals of this cyclic quadrilateral are perpendicular to each other.

DeleteCorrection - this will be only if ABCD is cyclic

DeleteLet ext∠A=2a, similarly for B, C, D.

ReplyDeletea+b+∠A₁=180°

c+d+∠C₁=180°

a+b+c+d+∠A₁+∠C₁=360°

But a+b+c+d=180°

Thus ∠A₁+∠C₁=180°

Hence A₁B₁C₁D₁ is concyclic.

Problem 1241

ReplyDeleteSet <A1AB=<D1AD=x,<A1BA=<B1BC=y,<B1CB=<DCC1=z and <CDC1=<ADD1=φ.Then

2x+2y+2z+2φ=180.But <AA1B+<DC1C=180-(x+y)+180-(z+φ)=180.Therefore A1, B1, C1

and D1 are concyclic.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE