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

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

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## Saturday, December 17, 2016

### Geometry Problem 1295 Right Triangle, Incenter, Incircle, Excenter, Excircle, Congruence, Angle

Labels:
angle,
congruence,
excenter,
incenter,
right triangle

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https://goo.gl/photos/cbTJjt1P5rtiULgY9

ReplyDeleteDraw points F, N, P, Q and M per attached sketch

Observe that BN=CP=CQ=IF= inradius of triangle ABC

Triangle CFI congruent to EQC ( case ASA)

So CI=CE => ICE is isoceles right triangle => CM/IE= ½

In right triangle ACM , CM/AC= CM/IE= ½

So ACM is 30-60-90 triangle

And angle A= 60 and angle C= 30

Draw from I parallel to AB, from E parallel to BC they meet at K

ReplyDeleteTr IEK congr to Tr ABC => ang EIK = ang C, but ang EIK = 1/2 ang A

=> ang C = 30

Problem 1295

ReplyDeleteIn-radius r = ½(c+a-b) and ex-radius ra = ½(b+c-a)

(ra+r)2 + (ra-r)2 = b2

2(ra2 + r2) = b2

(c+a-b)2 + (b+a-c)2 = 2b2

Which simplifies to a2+c2 =2bc

So b2 = 2bc and so b =2c

Hence ABC is a 30-60-90 triangle and <C = 30

Sumith Peiris

Moratuwa

Sri Lanka

Let M,N be the contact points of incircle, excircle respectively with AB; easily MN=BC, so BC is projection of AC onto BC, MN projection of IE onto AB. Since EI=AC, <A=2<C, thus <C=30.

ReplyDeleteBest regards