## Wednesday, August 18, 2010

### Problem 505: Right Triangle, Cevian, Sum of Segments, Angles

Geometry Problem
Click the figure below to see the complete problem 505 about Right Triangle, Cevian, Sum of Segments, Angles. Complete Problem 505

Level: High School, SAT Prep, College geometry

1. With the given data is it possible to prove that BD bisects /_B ? Easy to do this by trigonometry but how do we do it by plane geometry alone?
Of course, this makes x=15 deg.
Vihaan

2. draw DBM = x => BM = a, get AH = a + d , H on AC
=> CH = DM => HM = d => MA = a => MBA = 2x, B = 6x
=> x = 15°

1. https://photos.app.goo.gl/BkkquzcTCvWzbeUP6

3. To c.t.e.o

It not clear to me why CH=DM in your solution. Please explain.

Peter Tran

4. geting AH = a + d give us ang HBC = x
drawing DBM = x => ▲DMB = ▲HBC
(BM = BC = a, MCB = CMB, MBD = CBH = x : ASA )
In triangle MBC altitude is bisector, giving BM = a

5. To: c.t.o.e.
What a nice soluion, well done!
Ajit

6. To Joe
Thanks

If BG is altitude of ABC than BG is median of MBC,DBH
=> MD = HC as diferences of equal parts

( I liked your solution of P 167 )

7. To c.t.e.o
Your solution is the most beautiful solution I've ever seen.

8. Trisect < CBD by BE and BF, E and F on AC and BE thus being an altitude as well

Let CE = e so that DE = EF = d-e and CF = 2e-d

< AFB = 90-x = < ABF so AB = AF = a+d = b - (2e-d) hence e = (b-a)/2......(1)

Now CB is tangential to Tr. ABE at B so,

a^2 = eb

Substituting for e in (1),

b(b-a)/2 = a^2 hence b^2 - ab - 2b^2 = 0

Solving the quadratic and rejecting the negative solution we have b = 2a hence 2x = 30 and x = 15

Sumith Peiris
Moratuwa
Sri Lanka

9. This is my solution, enjoy it:

https://mega.nz/#!ZkwXWaRI!vn8pL5gHZ6hafXnKs0Y6TAgAyNAwUCcI-ECxJkwHtlI

Pedro Miranda