## Friday, June 13, 2014

### Geometry Problem 1021: Triangle, Cevian, Double Angle, Sides

Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 1021.

1. Draw CH prenpedicular to AB.
Let AH = x.
By pythagorean theorem AC^2 - AH^2 = CB^2 - BH^2 ----> B^2 - x^2 = (b + c/2)^ - ( c- x)^2
From short develop finally we get AH = x = 3c/8 - b/2
Now draw isoceles triangle CDP (continue AP ....)
Notice that:
1) <A = <CDP + <CPD = 2*<APC

1. Yes , sure but the solution is right......?

2. I mean DH = PH = 3c/4 - x = 3c/4 - 3c/8 +b/2 = 3c/8 + b/2
then DA = DH - x = 3c/8 + b/2 - 3c/8 + b/2 = b
from here CA = DA = b ----> <CDP + <CPD = 2*<APC = <A

3. I understand the idea of your solution.
Your step AH = x = 3c/8 - b/2 is OK
but your step: "I mean DH = PH = 3c/4 - x = ..."
should be: DH = AH then triangles AHC and DHC are congruent (SAS) then CD = b then ....

2. Does this problem have a purely geometrical proof? I proved it using trigonometry.

3. Applying Stewart's theorem,(cp=x)
b^2.c/4 + (b+c/2)^2.3c/4=c(x^2 +3c/4.c/4)
=>b^2+(b+c/2)^2.3=4(x^2+3c^2/16)
=>b^2+3b^2+3bc+3c^2/4=4x^2 +3c^2/4
=>b^2+3/4.bc=x^2 =>x^2=b(b+3c/4)
,',In ΔAPC, x^2=b(b+3c/4). By triangle properties, <A=2<APC

4. See my synthetic solution at my page https://www.facebook.com/photo.php?fbid=784793084934933&set=a.784793394934902.1073741826.100002127454113&type=1&theater

5. with C as center, radius b, make a circle O, circle O intercept AB at E, intercept CB at F;
Extend BC to intercept circle O at G.
CA=CE, /_ CAB = /_CEA=2x
so BA*BE=BF*BG; BA=c,let BE=y, BF=1/2c, BG=2b+1/2c;
so BF= b+1/4c,
so PE=b,
so trangle CEP is isoceles,
so /_ CAB =2 /_CPA

6. Extend PA to X such that XA = b. Let Y be the foot of the perpendicular from C to AP. Let AY = p.

Then b^2 - p^2 = (b+c/2)^2 - (c-p)^2 from which p = 3c/8 - b/2 and PY = 3c/8+b/2

So XY = PY and hence XC = CP and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

7. Solution 2

Mark Z on AP such that CZ = b
Then AZ = 2p = 3c/4 - b (from my previous proof) and so PZ = b = ZC
Hence < CZA = 2P = A

Sumith Peiris
Moratuwa
Sri Lanka