Proposed Problem

See also: Complete Problem 30, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Thursday, April 16, 2009

### Problem 30. Right Triangle, Incenter, Inradius

Labels:
incenter,
incircle,
inradius,
right triangle

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From F draw FD perpendicular to AB meeting the latter in D. Now ED = r, angle EAB = angle EAF = A/2, r = AE sin(A/2) and AE = AFcos(A/2) which gives us: r/sin(A/2) = AFcos(A/2) or AF = r/sin(A/2)cos(A/2) = 2r/sin(A). Finally, BG = FD = AFsin(A) = 2r/sin(A) * sin(A) = 2r.

ReplyDeleteAjit: ajitathle@gmail.com

Let G be an intersecting point of the segment AB and the extension of FE and let

ReplyDeleteP,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.

Then we see that BQ=EP=EQ=r and

two triangles EPG and ERF are congruence.

Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r.

Let G be an intersecting point of the segment AB and the extension of FE and let

ReplyDeleteP,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.

Then we see that BQ=EP=EQ=r and

two triangles EPG and ERF are congruence.

Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r

Through E draw MN ∥ AB with M on AF and N on BG.

ReplyDelete∠AEM = ∠EAB = A/2 -> ME = MA

∠MEF = 90° - A/2 = ∠MFE -> ME = MF

AB ∥ MEN ∥ FG

So BN = NG

But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°

-> EN = BN

So BN = NG = EN = r

Hence BG = 2r

(-> denotes "implies")

Slight Correction:

ReplyDeletein the line

"But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°"

it should be read as

" ∠BEN = ∠EBN = 45° "

Extend EF to cut AB at M; since M is the symmetrical of F about E and FG||AB, FG is tangent to the incircle and thus we are done.

ReplyDeleteBest regards,

Stan Fulger

Since FG is parallel to AB we can see that FE bisects < AFG. So E is the ex centre of Tr. FGC. Hence EG bisects right < BGF and so < EGB = 45. But < EBG = 45 and so BG= 2r where r is the in radius = altitude of isoceles Tr. BEG

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

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ReplyDelete