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.

Details: Click on the figure below.

## Tuesday, September 25, 2018

### Geometry Problem 1388: Triangle 40-60-80 degree, Incenter, Congruence

Labels:
40 degrees,
60 degrees,
80 degrees,
congruence,
geometry problem,
incenter,
triangle

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Make AD = AB, D on AC;

ReplyDeleteso ABD is equilateral triangle;

so angle DBC=20; angle BDC =120 = angle CIB,

so triangle BDC equal to triangle BIC;

so CI=BD=AB=c

Let D be the intersection point of the continuation of IC on AB, E be the intersection of the continuation of BI on AC and let x = IC and y = EC.

ReplyDelete1. tr ADC is similar to tr IEC so let k be the scale factor and CD = ky, AC = kx

2. Tr CBD is isosceles (40-80-80) so BC = CD = ky

3. Tr BEC is also isosceles (40-100-40) so EC = BE = y

4. Tr ABC is similar to AEC (40-60-80) so AB/AC = BE/BC = y/ky = 1/k

which implies AB = AC * 1/k = kx * 1/k = x or in other words AB = IC = x.

Considering usual triangle notations,

ReplyDeletelet BI intersect AC at D and CI intersect BA at E

We know AD=bc/a+c and DC=ab/a+c (Angle bisetor theorem)

Observe that BEC is isosceles and EC=a

Similarly BD is isosceles and BD=DC=ab/a+c --------(1)

Since BDA similar to CBA (AAA)

=> BD=CB.AB/AC

=> BD=a.c/b-----------(2)

(1)=(2)

=>c=b^2/(a+c) ------------(3)

Observe that DIC is similar to EAC

=>IC=AC.DC/EC

=>IC=b.(ab/a+c)/a (substitute values for DC and EC)

=>IC=b^2/(a+c) ------------(4)

From (3) & (4) the result follows

https://photos.app.goo.gl/M6AHCEaycVHP1j7v5

ReplyDelete