Geometry Problem 1507. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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According to the Diagram <B = 2@.....................Case 1

ReplyDeleteAccording to the Data given <B = @...................Case 2

Case 1:-

Let BE bisect <B, E on AC

Hence ABDE is concyclic and BE is perpendicular to AC

So AB = BC = 3+4 = 7

Case 2:-

<CAD = <B = @

So AC is tangential to circle ABD at A

Hence AC = sqrt (3 X 7) = sqrt21 and so

AD = sqrt (21 - 9) = sqrt12

Therefore AB = sqrt(12 + 16) = sqrt 28 = 2.sqrt7

Sumith Peiris

Moratuwa

Sri Lanka

Thank you, Sumith, for your solutions and observation. To unify the idea, the graph has been modified to address Case 1.

DeleteTr ABC => isosceles => AB=7 (the sketch will come soon)

ReplyDeletehttps://photos.app.goo.gl/QSZMRAiAmrBdUsbQA

Deletehttps://photos.app.goo.gl/v4gnknaopkZNweet5

Delete=7

ReplyDeleteThank you Peter for your prompt solution

ReplyDelete1st solution: Let CE be the C- altitude, H - orthocenter; BEHD being cyclic, <CHD=<B=2@, or <ACH=@, AB=BC.

ReplyDelete2nd solution: Take F reflection of C about D, AC is tangent to the circle AFB and from p.o.p. AC^2=CF.BC= 6.7 ; by Pithagoras AD^2=AC^2-BD^2=42-9=33 and AB^2=AD^2+BD^2=33+16=49, AB=BC=7