HELP!!!

Took me a while and I don't think I have the easiest method, but I've got an answer for question 2.

Let angle ABC (and all angles equal to it) be α

This allows us to make some deductions:

1)  angle AEB = 180°-α (supplementary to angle CED), and angle BAE = 180°-2α (sum of angles in isoceles triangle ABC), implies that angle ABE = 180°-(180°+α+180°-2α) = 3α-180° (sum of angles in triangle ABE), implies that angle CBD = α-(3α-180°) = 180°-2α (adjecent to angle ABE)

2) DE = 8-2 = 6 (isosceles triangle BCE)

3) if angle ABC = angle CDE and angle ACB = angle DCE, triangles ABC and CDE are similar

Using deduction 1, figure out the length of CD in terms of α using the cosine rule.

\(CD^2=BD^2+BC^2-2\times BD\times BC\times cos(CBD)\)

\(CD^2=2^2+8^2-2\times 2\times 8\times cos(180°-2α)\)

\(CD=\sqrt{68+32cos(2α)}\)

CD = EC (isoceles triangle CDE)

Using deduction 2, apply the cosine rule to triangle CDE to find a value for α (if you have a fear of long or ugly equations, look away now).

\(DE^2=CD^2+CE^2-2\times CD\times CE\times cos(DCE)\)

\(6^2=\sqrt{68+32cos(2α)}^2+\sqrt{68+32cos(2α)}^2-2\times \sqrt{68+32cos(2α)}\times \sqrt{68+32cos(2α)}\times cos(180°-2α)\)

\(0=100+64cos(2α)+200cos(2α)+64cos^2(2α)\)

Then solve the quadratic however you want (graph, formula, PlySmlt2)

\(cos(2α)=-0.625\)

\(α=64.341°\)

Use this value to find the length of CE and CD.

\(CD=\sqrt{68+32cos(2α)}\)

\(CD=\sqrt{68+32cos(2\times64.341°)}\)

\(CD=6.9282\)

Using deduction 3, find the length of AB with similar triangle rules.

\(\frac{DE}{BC}=\frac{EC}{AB}\)

\(\frac{6}{8}=\frac{6.9282}{AB}\)

\(AB=9.24\)

Feel free to ask if any of my thinking is unclear or I've made a mistake.

Since AT bisects <CAB

then    <CAT = <BAT = 30 degrees

<ABT=45 degrees

<ATC=75 degrees  exterior angle of a triangle = the sum of the 2 interior opposite angle.

<ACT =75 degrees  angle sum of triangle ACT

Now since <ACT=<ATC,  triangle ACT is isosceles and AC=AT=24 units

Now use sine rule to find AB

\(\frac{AB}{sin75}=\frac{24}{sin45}\\ AB=\frac{24sin75}{sin45}\\\)

AB= 24*sin(75)/sin(45) = 32.784609690803231

\(area=0.5*24*32.7846*sin(60)\)

0.5*24*32.7846*sin(60) = 340.7075574347631168

Area triangle ABC = 340.7 units^2        (1dp)

For the second question, begin by deducing the length of DE, and then, from the triangles DEC and BEC, calculate two expressions for the length of CE.

Equate them and that gets you the cosine of the common angle.

\(\text{Because we have isosceles triangle: } \\ \begin{array}{rcll} \text{Let } \overline{CE} = \overline{CD} = y \\ \text{Let } \overline{AB } = \overline{AC } = x \\ \overline{BC } = \overline{BE } = 8 \\\\ \overline{BD } = 2 \\ \overline{DE } = \overline{BE } - \overline{BD } = 8-2 = 6 \end{array}\)

\(\text{Let } \angle ABC = \angle ACB = \angle DEC = \angle CDE = \alpha \)

\(\text{Because we have a isosceles triangle: } \\ \begin{array}{rcll} \cos(\angle BCA) &=& \frac{ \frac{ \overline{BC} }{2} } { \overline{AC} }\\ \cos( \alpha) &=& \frac{ \frac{ 8 }{2} } { x }\\ \cos(\alpha) &=& \frac{ 4 } { x }\\ \mathbf{x} & \mathbf{=} & \mathbf{ \frac{4}{\cos(\alpha) } } \qquad | \qquad x = \overline{AB}\\ \end{array}\)

\(\begin{array}{lrcll} &&& \text{Because we have a isosceles triangle(BEC): } \\ (1) & \overline{CE}^2 &=& \overline{BE}^2 + \overline{BC}^2 - 2\cdot \overline{BE} \cdot \overline{BC} \cdot \cos(\angle EBC ) \\ & y^2 &=& 8^2 + 8^2 - 2\cdot 8 \cdot 8 \cdot \cos(180^{\circ}-2\alpha) \\ & y^2 &=& 128 - 128 \cdot \cos(180^{\circ}-2\alpha) \\ & y^2 &=& 128 + 128 \cdot \cos( 2\alpha) \\\\ &&& \text{In triangle(BDC): } \\ (2) & \overline{CD}^2 &=& \overline{BD}^2 + \overline{BC}^2 - 2\cdot \overline{BD} \cdot \overline{BC} \cdot \cos(\angle EBC ) \\ & y^2 &=& 2^2 + 8^2 - 2\cdot 2 \cdot 8 \cdot \cos(180^{\circ}-2\alpha) \\ & y^2 &=& 68 - 32 \cdot \cos(180^{\circ}-2\alpha) \\ & y^2 &=& 68 + 32 \cdot \cos( 2\alpha) \\\\ (1) = (2): & 128 + 128 \cdot \cos( 2\alpha) &=& 68 + 32 \cdot \cos( 2\alpha) \\ & ( 128-32) \cdot \cos( 2\alpha) &=& 68 -128 \\ & 96 \cdot \cos( 2\alpha) &=& -60 \\ & \cos( 2\alpha) &=& -\frac{60}{96} \\ & \cos( 2\alpha) &=& -\frac{5}{8} \qquad | \qquad \arccos() \\ & 2\alpha &=& \arccos(-\frac{5}{8}) \\ & 2\alpha &=& \arccos(-0.625) \\ & 2\alpha &=& 128.682187453^{\circ} \\ & \alpha &=& 64.3410937267^{\circ} \\ \end{array}\)

\(\begin{array}{lrcll} \mathbf{x} & \mathbf{=} & \mathbf{ \frac{4}{\cos(\alpha) } } \qquad | \qquad x = \overline{AB}\\\\ x &=& \frac{4}{\cos(64.3410937267^{\circ}) } \\\\ \mathbf{ x } & \mathbf{=} & \mathbf{ 9.23760430703 } \end{array}\)

\(\overline{AB} = 9.23760430703 \)

Very nice work to all involved, here !!!!

Here's another approach to (2)

Let the four equal angles in the question  = a

Since  < DEC = <  ACB  , then EB = BC

And EB = BD + DE   =  8    so....

8  = 2 + DE    →   DE  = 6

And <  ECD   = 180 - 2a

So....using the Law of Sines, we have :

sin DEC / DC  = sin ECD / 6

sin (a) / DC =  sin(180- 2a)/ 6

sin(a) / DC = sin(2a)/ 6

sin(a)/DC  = [2sin(a)cos(a)] / 6

DC /sin(a)  =  6 / [2sin(a)cos(a)]

DC = 6sin(a) / [ 2sin(a)cos(a)]

DC = 3sec(a)= EC

And using the  Law of Cosines, we have :

(DC)^2  = 2^2 + 8^2 - (2)(16)cos(DBC)

(3sec(a))^2  = 2^2 + 8^2 - (2)(16)cos(DBC)

cos(DBC) = [9sec^2(a) -  68] / [-32]      (1)

And applying it again, we have :

(EC)^2  = 2^2 + 8^2 - (2)(16)cos(DBC)

(3sec(a))^2  = 8^2 + 8^2 - 2(64)cos(DBC)

cos(DBC) =  [ 9sec(a)^2 - 128] / [-128]    (2)

Equate (1), (2)

[9sec^2(a) -  68] / [-32]  = [(9sec(a)^2 - 128] / [-128]

4[ 9sec^2(a) - 68]  = 9sec^2(a) - 128

36sec^2(a) - 272  = 9sec^2(a) - 128

27sec^2(a) = 144

sec^2(a)  = 144/27  =  16/3

cos^2(a)  = 3/16

cos(a)  = sqrt(3/16)

arccos(sqrt(3)/4) = a = 64.34°  = m< ACB

So m< BAC = [180 - 2a] = [180 - 2(64.34)]  ≈  51.32°

And using the Law of Sines once more, we have that :

AB/sin(ACB)  =   BC/ sin(BAC)

AB /sin(64.34) = 8/ sin(51.32}

AB = 8*sin(64.34)/sin(51.32)  = about 9.237

Thanks Will, Heureka and Chris,

You have all given great answers :)

I have another method for the second question.  I found the exact answer and used no trig.

As others have already shown, triangle BEC is isosceles so BE=BC=8

DE=8-2=6 units.

Let angle ECD= alpha

It follows that in triangle BCD

< BCD = theta-alpha

< CDB =180-theta

< DBC = alpha

Now consider triangle ABE

<ABE = theta-alpha

<BEA=180-theta

<EAB = alpha

so 

Triangle BCD is similar to triangle ABE       (3 equal angles)

Hence

\(\frac{AB}{BE}=\frac{BC}{DC}\\ \frac{AB}{8}=\frac{8}{DC}\\ AB=\frac{64}{DC}\\\)

Also triangle ABC and triangle CED are both isosceles triangles with base angles of theta degrees so they are also similar.

so

\(\frac{AB}{BC}=\frac{DC}{DE}\\ \frac{AB}{8}=\frac{DC}{6}\\ AB=\frac{4*DC}{3}\\~\\ so\\~\\ \frac{64}{DC}=\frac{4*DC}{3}\\ \frac{64*3}{4}=DC^2\\ DC=\frac{8*\sqrt3}{2}\\ DC=4\sqrt3\;\;units\\\)

\(AB=\frac{64}{DC}\\ AB=\frac{64}{4\sqrt3}\\ AB=\frac{16}{\sqrt3}\\ AB=\frac{16\sqrt3}{3}\;\;units\\~\\ AB \approx 9.2376\;\;units\)

Very  impressive, Melody      !!!!!!

Thanks Chris :)