## Sunday, March 20, 2016

### Geometry Problem 1201: Triangle, Interior Perpendicular Bisector, Midpoint, 90 Degrees, Area

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1201.

1. angle(BCA)=θ
triangle ABC ・・・cosθ=(a^2+b^2-c^2)/(2*a*b) ①
triangle DMC ・・・cosθ=(b/2)/DC=b/(2*DC)　　　②
①=②
DC=a*b^2/(a^2+b^2-c^2) ③

area　triangle DMC=S*(1/2)*(DC/a)=(b/2)*DM*(1/2)
∴DM=2*S*DC/(a*b)
=2*S*③/(a*b)=a*b^2/(a^2+b^2-c^2)

2. S = bh/2...(1) where h is the length of the altitude from B

From similar triangles

d/h = b/2/sqrt(a^2-h^2)...(2) where d = MD

Eliminating h between (1) and (2),

d = S/sqrt(a^2-4S^2/b^2) = 2bs / sqrt{4a^2b^2 - 16s(s-a)(s-b)s-c)} where s =(a+b+c)/2 which when simplifying the algebra using the difference of 2 squares yields

d = MD = 2bS/(a^2+b^2-c^2)

Sumith Peiris
Moratuwa
Sri Lanka

3. Draw altitude BH
Per law of cosine in triangle ABC we have Cos(C)= (a^2+b^2-c^2)/(2.a.b)
So CH= a. cos(C )= (a^2+b^2-c^2)/(2.b)
Triangles CBH similar to CDM so DM/BH= DM/DH= b^2/(a^2+b^2-c^2)
Replace BH= 2.S/b in above expression we get DM= 2.b.S/(a^2+b^2-c^2)

4. A simpler proof

Let H be the foot of the perpendicular from B to AC and let CH = e

Then applying Pythagoras to triangles ABH and CBH

c^2 - (b-e)^2 = a^2 - e^2 from which

e = ( a^2+b^2-c^2)/2b

From similar triangles

e/(b/2) = h/d where d = MD and h = BH

So (a^2+b^2-c^2)/b^2 = S/(b/2)/d

Hence MD = d = 2bS / (a^2+b^2-c^2)

Sumith Peiris
Moratuwa
Sri Lanka

5. S=1/2absinC=1/2abMD/DC=MD.a.MC/DC=MD.a.cosC=MD.a.(a^2+b^2-c^2)/2ab.

6. MD = MC tan C = (b/2) tan C = (b sin C)/(2 cos C) = (1/2) ab sin C)/(a cos C)
= S/a cos C) = 2bS/2ab cos C = 2bS/(a^2+b^2-c^2)

7. RHS=2abS/a(a^2+b^2-c^2)=S/a.cosC=ab.sinC/2a.cosC= tanC/(b/2)=tanC/MC=MD