## Saturday, February 21, 2009

### Problem 17: Right triangles and Angles

Proposed Problem

See complete Problem 17 at:
gogeometry.com/problem/problem017.htm

Level: High School, SAT Prep, College geometry

1. From Tr. ABC, AB/AC = sin(2θ)
AB = AC sin(2θ)=2ACsin(θ)cos(θ)
= 2AC * DE/DC * DC/AC from Tr. CED & ADC
Thus, AB = 2DE
Ajit: ajitathle@gmail.com

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3. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

4. Nilton Lapa
Take F on de extension of BC, so that CF=AC. Tr. ACF is isosceles, with angle(CAF)=angle(CFA)=(theta). Let BC be the median of ACF. Draw GH perpendicular to CF. DAC and GCF are congruent so GH=DE. As F is middle point of CF, we have GH=AB/2 and AB=2.DE.

5. AB/DE = AB/AC. AC/CD . CD/DE
= sin 2θ. sec θ. cosec θ = 2

6. Where I said "Let BC be the median of ACF", please change to "Let CG be ..."
Where I said "As F is middle point of CF", please change to "As G is middle poit of AF".
Nilton Lapa

8. Take F - reflection of A in CD abd G projection of F onto AC. Easily FG=2DE and triangles CFG and CAB are congruent, hence FG=AB, done.

Best regards,
Stan Fulger

9. Let AC, DB meet at F and let AG drawn parallel to BC meet BD at G.

ABCD is con cyclic and Tr.s ADF, BCF & AFG are all isoceles.

Now Tr.s DEF & BAG being similar AB/DE = AG/EF = 2. AG/AF = 2 since AG=AF

Hence AB = 2.DE

Sumith Peiris
Moratuwa
Sri Lanka

11. Another idea: take F reflection of D in E; AFBCD is cyclic and we need DF=AB. Since CF is angle bisector of <ACB, there exists the equality BF=AF; but as constructed AF=AD, thus DAFB is an isosceles trapezoid, having equal diagonals, DF=AB, done.

12. Solution 2

Following on from my previous solution

Let AE = EF = p and let CF = BC = q
AB^2 = (2p+q)^2 - q^2 = 4p(p+q) = 4DE^2 and the result follows

Sumith Peiris
Moratuwa
Sri Lanka