## Thursday, December 3, 2020

### Geometry Problem 1489: Right Triangle, Angle Bisectors, Perpendicular, Measurement

Geometry Problem 1489. 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/wzBcnmF6f4N9989D6

Draw incircle D with radius= r
Define position of M, N and P ( see sketch)
Let ∠ (A)=2u and ∠ (C)=2v
We have ∠ (DEF)=180-∠ (DEA)- ∠ (FEC)= u + 2v
Replace 2v= 90- 2u we get ∠ (DEF)=90-u=∠ (DEA)
Triangle DME congruent to DPE ( case HA) => EF tangent to incircle D
So DPFN is a square and DF= r. sqrt(2)
But r=2 ( see problem 1487) so DF=2.sqrt(2)

2. Inradius of Tr. AB, r = S/s = (48/2)/12 = 2

If U is the tangent point of the incircle on AC, AU = s - a = 12 - 8 = 4 and so
From similar Tr.s AUD & DUE, UE = 1, so E is the mid point of AC & F is the mid point of BC (mid point theorem)

Hence BF = 4

If V is the point of tangency on BC of the incircle, BV = s-b = 12-10 = 2
Hence VF = 4-2 = 2

So Tr.s BVD & FVD are congruent SAS being also right isosceles and DF = BD = 2.sqrt2

Sumith Peiris
Moratuwa
Sri Lanka

3. Extend AD to P, P on BC. DPFE cyclic => Tr BDF rightisosceles
=> DF=DB=2V2

1. https://photos.app.goo.gl/TZto12GVkGKjP58p6

4. Draw perpendicular from D to meet AC at P, BC at Q. DP=DQ=radius of incircle = 2
Extend AD to meet BC at O. Since AO is Angle bisector BO/OC=3/5=> BO=3,OC=5
=> PE=1, AP=4 => EC =5 and E is mid-point of AC and since EF||AB=>FC=4
Hence OF=1. Since BQ=2=> QO=1
So DQF is isosceles right triangle where DQ=QF=2=> DF=2.Sqrt(2)