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.

Click the figure below to view more details of problem 1236.

## Tuesday, July 12, 2016

### Geometry Problem 1236: Intersecting Circles, Secant, Concyclic Points

Labels:
concyclic,
intersecting circles,
secant

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Isosceles triangles PEA and QDA are similar => EA/PA = DA/QA => EA.AQ = PA.AD

ReplyDeleteHence E,P,Q and D are concyclic.

B also lies on the same circle

"B also lies on the same circle"

DeleteCould u pls explain how u arrived at this conclusion?

<PAQ = <PBQ = 135 which are supplimentary. Hence B also lies on the same circle. For some reasons, this did not get printed in my first comment. Faced this trouble couple of times and I stopped posting. ( I have posted <PEA = <PEQ = 45 in the beginning, but the preview is truncating this as well)

Delete< PAQ = < PBQ = 135. How did u arrive at this conclusion?

Deleteconnect PB, BQ, <PBQ = <PAQ,

ReplyDelete<PAQ + <PAE = 180 = <PAQ + <PEA =< PBQ + <PEA =180,

SO b ALSO LIES ON THE SAME CIRCLE

Isosceles triangles PEA and QDA. Since <PBQ+<E=<PAQ+<EAP=180, E,P,B,Q concyclic. Similary, D,Q,B,P concyclic. Therefore, E,P,B,Q,D concyclic.

ReplyDeleteProblem 1236

ReplyDeleteIs <EPA=2.<EBA (PE=PA), <AQD=2.<ABD (AQ=QD).But <EAP=<DAQ=x. Then <EPD=<EPA=

180-2x=2(90-x)=<AQD=<EQD.But <EBD=<EBA+<ABD=(< EPA)/2+(<AQD)/2=90-x+90-x=180-2x.Therefore points E,P,B,Q and D are concyclic.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

The solution that I propose is the following:

ReplyDelete<PEA=<ADQ=<EAQ=<DAQ=x ; <EPA=AQD=y= 180-2x

<PAB=<PBA=α ; <APB=K=180-2α

<QAB=<ABQ=β ;<AQB=R=180-2β ; <ABQ+<ABP = α + β

To be concyclic the points BPED and Q need that:

<R+<K+<y=180 or α + β +x= 180 ;

<R+<K+<y=180-2β+180-2α+180-2x=α + β +x=180

Therefore point BPED and Q are concyclic

Reply

https://goo.gl/photos/1g61g7m5f8PYXRTeA

ReplyDeleteConnect PQ, EB, BD, PB and QB ( see sketch)

Note that u= ∠ (BEQ)= ∠ (APQ)= ∠ (BPQ)

So E, P, B and Q are cocyclic.

Similarly P, B, Q and D are cocyclic