Need a hint with a geometry problem (ABC is isosceles; find MH_3)

[Tarrok]

Hello!

Could you guys give me some hints on how to solve the following problem (sketched in the link below):
(Sorry, I could not attach the image directly since the file is too large)

< link to objectionable page removed >

I know that the triangle ABC (AC=BC) is an isosceles one, but that doesn't help much.
Thank you for help.

 

[Tarrok]

You made your 1st post Oct.24th: did not respond.

This is your 2nd post.
WHERE are you stuck?
Remember: we do not do homework.

Thank you for the response.

I eventually solved the problem in my first post and responded in the topic that it was solved, but it seems my post didn't get through for some reason.

As for this one, in the picture below I marked everything I myself can find in this triangle.

< link to objectionable page removed >

I feel like what I should be looking at is the triangle AMB, where MH_3 is the height, but I don't know anything in this triangle.
Another idea would be to write down a system of equations for MH_3 using the law of cosines in the triangles MH_1/H_3 and MH_2/H_3. However, this approach also doesn't achieve much for the very same reason.
Finally, there might be a useful pair or two of similar triangles here, but if there are any, I can't find them.

 

[Deleted member 4993]

CH₁MH₂ is a cyclic quadrilateral.

What are the relevant theorems?

 

[hoosie]

Note:
Triangle ACB is isosceles with sides CA and CB each of length 20,
base AB and height CH3.


Method 1:


CA = 20
CB = 20
CH1 = 12
CH2 = 12
H1A = 8
H2B = 8


Using similar triangles:


CH1/CA = MH1/OA
12/20 = MH1/8
12*8 = 20*MH1
MH1 = 96/20
= 4.8
By symmetry:
MH2 = 4.8


(CO)^2 = 20^2 + 8^2
CO = (400 + 64)^(1/2)
= 464^(1/2)
= 21.5407


Using similar triangles:


CO/CM = CA/CH1
21.5407/CM = 20/12
21.5407 * 12 = 20 * CM
CM = 12.9244


Use trigonometry to determine (angOCB):


tan(angOCB)
= OB/CB
= 8/20
= 0.4
=> angle OCB = 21deg 48min



MH3 = CH3 - CM
cos(21deg 48min)
= CH3/20
CH3
= 20 * 0.928477
= 18.5695
MH3
= 18.5695 - 12.9244
= 5.6451
= 5.645 rounded to 3 decimal places





Method 2:


tan(angOCB)
= 8/20
= 0.4
angle OCB = 21deg 48min
21deg 48min + angle COB = 90deg
=> angle COB = 68deg 12min
cos(68deg 12min) = OH3/8
OH3
= 8 * 0.3714
= 2.9711

By Pythagoras theorem:
(CM)^2
= (CH1)^2 + (MH1)^2
= 12^2 + 4.8^2
= 167.04
CM
= (167.04)^(1/2)
= 12.9244


CM + MH3 + OH3 = CO
12.9244 + MH3 + 2.9711 = 21.5407
MH3 + 15.8955 = 21.5407


MH3
= 21.5407 - 15.8955
= 5.6452
= 5.645 rounded to 3 decimal places