Geometry Problem 1483. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Wednesday, November 11, 2020

### Geometry Problem 1483: Isosceles Right Triangle, Excenter, Perpendicular, Measurement

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https://photos.app.goo.gl/h5V695JxQirG1RF98

ReplyDeleteDefine point F per sketch

We have ∠ (ADF)= ∠ (ADB)= pi/8

FD= 15*cos(pi/8)

BD=FD*sqrt(2)

EB=BD*sin(pi/8)

So EB=15/2*sqrt(2)*sin(pi/4)

EB=15/2

https://photos.app.goo.gl/mVTbQNZAG1vg6GvU9

DeleteGeometry solution: Draw line DFP perpendicular to BA ( see sketch)

Note that triangle BDP is isosceles and BP= 2.BE

Triangle BDF is isosceles right triangle

Triangle BFP congruent to DFA ( case ASA) so

AD=BP=15 so BE= ½ BP= 7.5

Join D to B, D to C. Draw DH perpend to BC, DK perpend to ED.

ReplyDelete=> Tr DBK isosceles. Draw BM perpend to DK => BM median

=> DM = EB = 7.5

https://photos.app.goo.gl/6vkuEJhh9Tik5LcM8

DeleteComplete square BXDY. Let BD, AC cut at Z.

ReplyDeleteTr.s ADX, ADZ, ZDC & CDY are all congruent and Tr. ADX is also congruent with BXZ, ASA

So BE = BZ/2 = AD/2 = 7.5

Sumith Peiris

Moratuwa

Sri Lanka