The Right Brothers

Using that the perpendicular from D in AB give us the point E and the perpendicular from D in AC give us the point F, we can use the fact that the angle between AB and BC are equal to the angle between AC and BC (Isosceles Triangle) and we will call it ∂. Then, since we know that BD+DC=60, we can discover BD and DC using that sen(∂)=FD/DC=ED/DB.

As were told, 16=ED and 32=FD. So 32 DB=16 DC, or simple 2*DB=DC. Then we know that DB=20 and DC=40.

Knowing that we use that cos(∂)=FC/DC=(BC/2)/AC=BC/(2*AC) To discover FC we use Pitagora's theorem, using DC=40 and FD=32 we discover that FCˆ2=40ˆ2-32ˆ2 or FC=24

Then, cos(∂)=5/3 and 2 AC 3=BC*5 and BC=60. Then AC=50. And since AC=AB, AB=50.

$\triangle ABC=\triangle ADB+\triangle ADC$ $\frac{a\,\sqrt{4\,{b}^{2}-{a}^{2}}}{4}=\frac {1}{2}16\,c+\frac {1}{2}32\,b$ $\frac{a\,\sqrt{4\,{b}^{2}-{a}^{2}}}{4}=\frac {1}{2}16\,b+\frac {1}{2}32\,b$ $\frac{a\,\sqrt{4\,{b}^{2}-{a}^{2}}}{4}=24\,b$ $\frac{60\,\sqrt{4\,{b}^{2}-{60}^{2}}}{4}=24\,b$ $5\,\sqrt{4\,{b}^{2}-3600}=8\,b$ $\sqrt{4\,{b}^{2}-3600}=1.6\,b$ $4\,{b}^{2}-3600=2.56\,{b}^{2}$ $1.44\,{b}^{2}=3600$ ${b}^{2}=2500$ $b=50$

Call the intersection of the perpendiculars and $AB$ and $AC$ $E$ and $F$ , respectively. Since angles B and C are congruent and triangles $BED$ and $CFD$ are right triangles, they are similar. Then $BD$ and $DC$ split $BC$ into a 1:2 ratio, so $BD=20$ and $DC=40$ . Applying the Pythagoream Theorem to (BED) and $CFD$ gives us $12$ and $24$ , respectively.

Now draw $AD$ . This gives us 2 more right trangles $ADE$ and $ADF$ . Since they share the same hypotenuse, we know that the sum of the squares of their sides are equal; let $AE=a$ and $AF=b$ .

$a^2+16^2=b^2+32^2$

$a^2=768+b^2$

Since $AB=AC$ , we also have

$a+12=b+24$ .

$a-12=b$

Solving the system of two equation by substituting this into $a^2=768+b^2$ gives us $a=38$ , so $AB=12+a=\boxed{50}$ .

As with any geometry problem, we must draw what we have: $\bigtriangleup ABC$ with $AB=AC$ , point $A$ as the vertex, and $BC=60$ .

Then, we must assign variables to the triangle. We know that $AB=AC$ , so let us assign a variable $z$ to that side length. Moreover, we know that $D$ is $16$ units and $32$ units away, respectively, from $AB$ and $AC$ perpendicularly meaning that it meets the edges, respectively, at $B'$ and $C'$ at right angles, allowing us to make right triangles.

Now we have two right triangles that share portions of $BC$ has hypotenuses: (1) the one with hypotenuse $BD$ , a leg $B'D$ of length $16$ , and a leg $B'B$ of length we shall call $x$ and (2) the one with a hypotenuse $CD$ , a leg $C'D$ of length $32$ , and a leg $C'C$ of length we shall call $y$ .

In addition to the above right triangles, we have two more who share $AD$ as a hypotenuse: (1) $AB'D$ and (2) $AC'D$ . We can see that $AB'=z-x$ and $B'D=16$ , and that $AC'=z-y$ and $C'D=32$ .

Back to the first pair of right triangles. A trained geometer's eye will note that $DBB'$ and $DCC'$ are similar triangles by Angle-Angle-Angle (AAA) Similarity. $\angle B'BD = \angle C'CD$ because of the fact that $AB=AC$ . Moreover, $\angle AB'D = \angle AC'D = 90 ^ \circ$ because of the fact that they are right angles formed by perpendicular intersections. Still, $\angle B'DB = \angle C'DC$ because the other two angles have already been proven equal. Thus, $\bigtriangleup DBB' \simeq \bigtriangleup DCC'$ .

Because they are similar, it follows that $\frac {B'D}{C'D} = \frac {BD}{DC} \Rightarrow \frac {B'D}{C'D} = \frac {BD}{60-BD} \Rightarrow$ $\frac {16}{32} = \frac{1}{2} = \frac {BD}{60-BD} \Rightarrow 60-BD=2BD \Rightarrow BD=20$ and $CD=60-20=40$ .

We can now solve for $x$ and $y$ by using Pythagorean Theorem. $16^2+x^2=20^2 \Rightarrow 256+x^2=400 \Rightarrow x^2=144 \Rightarrow x=12$ . $32^2+y^2=40^2 \Rightarrow 1024+y^2=1600 \Rightarrow y^2=576 \Rightarrow y=24$ (or you could have just multiplied $x$ by $2$ because of similarity scaling.)

Finally, in order to solve for $z$ , we must "solve" for $AD$ . We can do this by using Pythagorean Theorem again. $(z-12)^2+16^2=(z-24)^2+32^2 \Rightarrow$ $z^2-24z+144+256=z^2-48z+576+1024 \Rightarrow$ $z^2-24z+400=z^2-48z+1600 \Rightarrow$ $-24z+400=-48z+1600 \Rightarrow z=50=AB$ .

$\star Kwesi L.$

Lets assume --

1. The perpendicular from D meets AB At Q and AC at P .

2.AP = x .and CP = y . 3.AQ= a and BQ=b .

4.BD= l and CD = m.

Therefore , x+y = a+b .......... (1)

l+m = 60 ...........(2)

16^2 + b^2 = l^2 ..(3)

32^2 + y^2 = m^2 ...(4)

Now , let us Look at triangles ,DQP and DCP .....

We conlcude that ,

\angle DQB = \angle DPC = 90 ^ \circ

In triangle ABC , AB= AC ...therefore , opp angles to equal sides are equal ...

Therefore . \angle QBD = \angle PCD

By Angle Sum property of a triangle , We say that \angle QDB = \angle PDC .

Therefore by AAA similarity , triangle DAB is similar to triangle DPC .

but , PD = 2QD ...

We can conclude that triangle DAB = twice triangle DPC .

y = 2b .... (5)

m =2l ...... (6)

Now , On BC ,

l + 2l = 60

therefore , l =20 ! ..... (7)

So ...

32^2 + (2b)^2 = (2l^2)

1024 + 4b^2 = 4l^2

256 + b^2 = l^2

from (7) , b^2 = 400 - 256

b = 12 ...... (8)

now , CP = 2QB ... From AAA similarity , ..... (9)

In quadrilateral AQDP , Let us name AD as r ....

a^2 + 16^2 = r^2

x^2 + 32^2 =r^2

therefore , Equating Above ones , we get ...

a^2 + 16^2 = x^2 + 32^2

a^2 - x^2 = 32^2 - 16^2

(a+x ) (a-x ) = 768 ...

a+x = 64 .... (10)

NOW ,

x+24 = a+12

12 = a - x

64 = a + x

76 = 2a

a = 38 ! ... (11)

From (10 )

a + x =64

38 + x = 64

x= 26 ...... (12)

                           So ... In AC , x + 24 = AC 

                                  AC = 24 + 26 = 50 ...

Since AC = AB .....

AB = 50 !!!

Let the intersections of the perpendiculars from $D$ to $AB$ and $AC$ be $E$ and $F$ respectively. Then, $\triangle DEB$ is similar to $\triangle DFC$ with scale ratio $1:2$ , so $BD=20$ and $DC=40$ . Then, by Pythagoras $BE=12$ and $CF=24$ . Let $EA=x$ . Then by Pythagoras $DA=\sqrt{x^2+{16}^2}$ and $FA=\sqrt{x^2-768}$ , so $12+x=24+\sqrt{x^2-768}$ . Solving this gives a single solution $x=38$ , so $AB=12+x=50$ .

let AB = x ,and the altitude of triangle from A=t we know that area of triangle ABC = area of triangle ABD + area of triangle ACD

so
(60×t)/2=(x ×16)/2+(x × 32)/2 and 30t = 24 x

       t=4/5 x

we also know that x^2 = t^2 + (60/2)^2 x^2 = 16/25 (x^2) + 900 9/25 ( x^2)= 900 x^2 =2500 x=50

Let $BD = x, DC = 60-x$ Call $\angle B = \angle C = \theta$ . We equate the sines of the two angles to give us $EB = 12, FC = 24$

Next \( \angle A = \pi - \theta) and using the sine rule we get

\( \frac{sin(\pi - \theta)}{BC} = \frac{sin(\theta)}{AC} \)

$\Rightarrow AC = \frac{30}{cos(\theta)}$ where $cos(\theta) = \frac{12}{20}$

$\Rightarrow AC = 50 = AB$

Assume the length of AB is t . Then AB = AC = t Area of triangle ABD = \frac {1}{2} .16 t = 8 t Area of triangle ACD = \frac {1}{2}.32 t = 16 t If we sum the area of both triangle ABD and ACD , we'll get area of triangle ABC , that is 24 t .

Look at triangle ABC . Let's draw an altitude line from A to line BC . We get the length of the altitude from phytagorean theorem, that is \sqrt{ AB ^2 - 30^2} = \sqrt{ t ^2 - 900}

Then we also get area of triangle ABC , \frac {1}{2} . 60 . \sqrt{ t ^2 - 900} = 30 t . \sqrt{ t ^2 - 900}

from the above, 24 t = 30 t . \sqrt{ t ^2 - 900}

Then, we'll get t = 50

So, the length of AB is 50.

Let DM is perpendicular to AB and DN to AC. Clearly, Triangle BDM ~ CDN (all angles are equal). => BD:DC = MD:DN => BD = 20 => Angle B = 57 degrees

Apply cosine formula in triangle ABC at angle B.

It is given that triangle ABC is isosceles with AB=AC, Thus, <B and <C are equal, Let <B=<C=X(in degrees) and <A=180-2X

Also,D is a point on BC, where BC=60 let CD=y and BD=60-y

sinX=32/y=16/(60-y) solving for y, we get y=40 Thus, sinx=4/5

Using cosine law, BC^2=AB^2+AC^2-2 AB AC*cos(180-2X) 3600=2AB^2+2(AB^2)(cos2X) 3600=2AB^2(1+1-2sin^2 x) 3600=2AB^2(2-2(16/25)) 3600=(36/25)AB^2 AB=50

If <BAC = x => <ACB=<CBA= 90-(x/2) ; => <EDC=<FDB=x/2 ; (E and F are points at which perpendicular from D meets AC and AB respectively) ; =>cos(x/2)=16/BD=32/CD ; =>BD=2CD ; =>BD=20, CD=40 (BD+CD = BC = 60) ; =>Cos(x/2)=4/5 => Cos(x/2)=3/5 ; Also, if we draw a perpendicular from A to BC it would meet at the midpoint G of BC ; Hence, sin(x/2) = BG/AB (BG=BC/2=30) ; => AB=BG/sin(x/2)=30/(3/5)=50 ;