Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1261.

## Friday, September 16, 2016

### Geometry Problem 1261: Cyclic Quadrilateral, Circle, Diameter, Midpoint, Measurement

Labels:
circle,
cyclic quadrilateral,
diameter,
measurement,
midpoint

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https://goo.gl/photos/tDgKutmrL4Bn9Lur8

ReplyDeleteDraw points O, E, F, G per sketch

Note that OE ⊥AB , AF⊥ EF and triangles EBC, AFE are 45-45- 90 triangles

We also have AF/FC= CD/AD= 1/3 => EC//AE

And BE/BA=BC/BE= 1/2

Dear Peter - How is AF/FC = CD/AD?

DeleteSince AB=2BC=> BC=BE=AE ( see sketch)

DeleteTriangles AFE, EGB and BGC are 45-45-90 triangles and are congruent ( case ASA)

so FA=FE=EG=GC => AF/FC= 1/3

Palai Renato

ReplyDeleteWe put AB=1, BC = 2 AC= √5,

We put CD=x we have x+(3x)2=√5, and therefore x= 1/√2

<BAC =45 (regle of sum of tangent), CD=DE and CE=1

CE=CB and C is midpoint of BE

Reply

How is < BAC = 45?

ReplyDeleteABCD is cyclic, with AC being diameter. Hence, Angle ABC and Angle ADC are right angles, by Thales' theorem.

ReplyDeleteBy pythagoras’ theorem, it’s evident that BC = AC/sqrt(5), AB = 2AC/sqrt(5), CD = AC/sqrt(10), AD = 3AC/sqrt(10). We need the sine and cosines of angle BAC, angle CAD.

If we let O be the circumcenter of ABCD, then angle BOC = 2 angle BAC, angle COD = 2 angle CAD. This is obvious high school geometry.

From here, we use complex numbers.

Let the circle on ABCD be the unit circle and c = 1, a = -1.

With a bit of trig (especially double angle formula), b = cis(angle BOC) = cis(2 angle BAC) = (3/5) + i(4/5). Similarly, d = cis(-angle COD) = (4/5) – i(3/5).

Now let’s make a point F such that C is the midpoint of B and F. This indicates (f + b)/2 = c, which rearranges to f = 2c – b. We compute f = (4/5) – i(4/5).

We now assert that A, D and F are collinear. This is proved by checking that (a – f)/(a – d) is a real number (justification is De Moivre theorem; (a - d) multiplied by a real number would make (a-f) iff a-f and a-d make an angle of 0 or 180 degrees with each other).

(a – f)/(a – d) = (12/5 – i(4/5))/((-9/5) – i(3/5)) = (4/5) / (-3/5) = -4/3 which is definitely a real number. So we conclude that F is on line AD and BC (as C is midpoint of BF).

But we know that E is the intersection of AD and BC. This means E = F, and the proof is complete.

Problem 1261

ReplyDeleteIs <BAC+<CAD=45, so <CED=45 then CD=DE. Let the points K, L on the side AD such that

AK=KL=LD=AD/3=DC=DE. So AL=LE and BL is perpendicular at AE so BL//CD. Therefore

BC=CE (LD=DE).

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Palai Renato,

ReplyDeleteReply to Sumith Peiris to request to motivate why angle.. is 45°.

If we calla <BAC= and <CAD= , <BAD= + , bat

tang ( + )=(tang + tang )/(1 - tang tang ) and introducing the values of and , we have: (2+3)//1-2x3)= -1, tang 1 = 45°

Reply

Thank u. Good pure geometry proof

ReplyDeleteGood pure geometry solution

ReplyDeleteLet CD = 1

ReplyDelete=> AD = 3

=> AC^2 = CD^2 + AD^2 = Sqrt(10)

Similarly solving for AB & AC in the right triangle ABC,

we get BC = sqrt(2) --------(1)

and AB = 2*Sqrt(2)

Let CE = x and DE = y

As CE^2 = DE^2 + CD^2

=> 1+y^2 = x^2 ----------- (2)

As ED.EA = EC.EB (From Secant-Secant Rule)

=> y.(3+y) = x.(x+Sqrt(2))

=> y^2+3y = x^2+Sqrt(2).x

=> 3y = (x^2-y^2) + Sqrt(2).x

=> 3y = 1 + Sqrt(2).x (from (1) )

=> y = [1+Sqrt(2).x]/3 ----------(3)

Substituting the value of y in Eq(2)

=> 9 + (1+2x+2Sqrt(2)x) = 9x^2

=> 7x^2 - 2Sqrt(2)x -10 = 0

Therefore x = (2Sqrt(2) + Sqrt(8+288))/14 = 14.Sqrt(2)/14 = Sqrt(2) = BC (from (1) )

Hence C is midpoint of BE