## Wednesday, November 11, 2020

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

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

Define point F per sketch
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

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

2. Join D to B, D to C. Draw DH perpend to BC, DK perpend to ED.
=> Tr DBK isosceles. Draw BM perpend to DK => BM median
=> DM = EB = 7.5

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

3. Complete square BXDY. Let BD, AC cut at Z.

Tr.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