Geometry Problem 1492. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Wednesday, December 30, 2020

### Geometry Problem 1492: Right Triangle, Altitude, Incenters, Angle, Measurement

Labels:
altitude,
angle,
geometry problem,
incenter,
incircle,
measurement,
perpendicular,
right triangle

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https://photos.app.goo.gl/tKctKLGSK736Hust7

ReplyDeletelet ∠ (DO2O1)= x

note that C, O and O2 are colinear and ∠ (CDO2)= 45

Triangle ADB and BDC are similar

So DO2/BC=DO1/AB => triangles ABC and O1DO2 are similar ( case SAS)

So ∠ (DO2O1)= ∠ (ACB)

In triangle DO2C, external angle ∠ (OO2D)= x+27=45+x/2

So x= 36 degrees

Extend O1O2 to P. From Tr O2PC => Ang C=36, From AngO1O2P = ?+117+27

ReplyDelete=> ?=36

https://photos.app.goo.gl/iojeJUo4Q3k2kqeC6

DeleteTR.s AO1D & BO2D are similar having angles 45, A/2, 135 -A/2

ReplyDeleteSo O1D/AD = O2D/BD and so O1D/O2D = AD/BD = (c^2/b)/(ac/b) = c/a

It follows that Tr.s O1DO2 & ABC are similar since < O1DO2 = 90 = <B

Therefore ? = <C and since C/2 + 45 = C + 27,

? = C = 36

Sumith Peiris

Moratuwa

Sri Lanka

Join BO and observe that Triangle BOD is similar to CO2D and m(O1DO2)=90

ReplyDeleteHence BD/DC=O1D/O2D

but BD/DC=AB/BC => O1DO2 is similar to ABC with m(DO2O1)=C

27+C=45+C/2

=>C=36

Correction: Join BO1 and triangle BO1D is similar to CO2D

DeleteCan we expect more geometry problems soon?

DeleteBest wishes Jens Carstensen, Frederiksberg, Denmark