## Wednesday, September 30, 2009

### Problem 361: Right triangle, Incircle, Incenter, Tangency points, Angle

Proposed Problem
Click the figure below to see the complete problem 361 about Right triangle, Incircle, Incenter, Tangency points, Angle.

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Complete Problem 361
Level: High School, SAT Prep, College geometry

#### 7 comments:

1. We've angle AED = angle AED = (180-A)/2 = 90 -A/2
Futher from Tr. FEC, α = angle AED - angle ACF =90-A/2-C/2 = angle B/2 since angs.(A+B+C)=180 deg. which, in turn, means that α = 90/2 = 45 deg.
Vihaan: vihaanup@gmail.com

2. @vihaan
Pls explain this step

We've angle AED = angle AED = (180-A)/2 = 90 -A/2

How do we get this?????

3. /_AED+/_ADE+/_A=180 but /_AED=/_ADE (equal tangents). So 2/_ADE+/_A=180 or /_ADE=/_AED=90- A/2
Vihaan

4. Since Tr. ADE is isoceles and OE is perp. to AC < DEO = A/2. But < FCE = 45-A/2 and hence considering the angles of Tr. FEC which add upto 180 we see that alpha= 45

Sumith Peiris
Moratuwa
Sri Lanka

5. Problem 361
The ADOE is cyclic (<ADO=90=<AEO) so <EDO=<A/2.Now DO//BC then <FOD=<FCB=<C/2.
But x=<EFO=<FDO+<DOF=<A/2+<C/2=45.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

6. Draw in the tangents lines and then just angle chase to find <OFE.

First let G be the intersection of the tangent on BC and x = <OCB

1. Tr. CGO = Tr.GEO by SSS. So <ECO also equals x.
2. <COE is 90 -x and <EOF is 90 + x.
3. DO is parallel to BC so <DOF is also x.
4. That means <DOE is 90 + 2x and since tr. EDO is isosceles (2 radii) <EDO - <DE0 = 45 - x.
5. Finally <EFO = 180 - (45 - x + 90 + x) = 45.

7. Considering usual triangle notations,Extend CO to meet AB at P
Since ADE is isosceles, m(ADE) = m(ADF) = 90-A/2
Also m(OCB) = 45-A/2 => m(CPB) = 45+A/2
Therefore m(PFD) = m(EFO) = 180-(90-A/2)-(45+A/2) = 45