Distance Between Perpendicular Feet

Remark that $AECF$ is cyclic because $\angle AEC = \angle AFC = 90$ . Now, as $ABC$ is isosceles the altitude $AE$ bisects $AE$ so it follows $EC = 130$ . Now by Pythagorean Theorem we get $AE = 312$ . I claim that $AE = EF$ . To prove this it suffices to show $\angle EAF = \angle EFA$ . To prove this, just observe that using $AECF$ is cyclic we have $\angle EAF = 180 - \angle ECF = 180 - \angle BCD = \angle ABC = \angle ACE = \angle EFA$ and thus $EF = 312$ as desired.

Since CAB is isosceles, E is the middle point of BC. AEB is a right-angled triangle, and AE = $\sqrt{AB^2 - EB^2} = \sqrt{338^2 - 130^2} = 312$ . Draw a line perpendicular to AB and CD through E. It crosses AB in a point G and CD in a point H. The triangles EGB and EHC are congruent, and therefore E is the middle point of GH. The triangles FEH and AEG are congruent, and so EF = AE = 312.

We claim that $EA=EF=312$ .

Since $\angle AEC=\angle AFC=90^{\circ}$ , we know that quadrilateral $AECF$ is cyclic. Also that since $AB=AC=338,$ $\angle ABC=\angle ACB$ , and since $ABCD$ is parallelogram, $\angle ABC+\angle BCF=180^{\circ}$ . Hence $\angle EAF=180^{\circ}-\angle ECF=\angle ABC=\angle ACB=\angle ACE=\angle AFE.$ Therefore, $EA=EF$ .

To find $EA,$ since $AB=AC, AE\perp BC$ , $AE$ is perpendicular bisector of $BC$ , and thus $E$ is midpoint of $BC$ . So $BE=EC=\frac {BC}{2}=\frac {260}{2}=130.$ Therefore, since $\angle AEC=90^{\circ}$ , by Pythagoras' theorem, $AE^{2}=AC^{2}-EC^{2}=338^{2}-130^{2}=13^{2}(26^{2}-10^{2})=13^{2}(576)=13^{2}\cdot 24^{2}=312^{2}$ . Hence, $AE=EF=312.$ Q.E.D.

We will try to find EF using law of cosines on ∆ECF. To do this we will need to find $EC, CF,$ and $cos\angleECF.$

$EC = BC/2 = 130$ because ∆ABC is isosceles.

$CF^2 = AC^2 - AF^2$ by the Pythagorean theorem. Using the area of ∆ACD, we know that $AE * AD = AF * DC.$ Then $260(312) = 338AF, so AF = 240.$

Using the Pythagorean Theorem on ∆ACF, we get that CF = 238.

Looking at $\angle ECF, \angle ECF = 90^\circ + \angle ABE.$ Let $\angle ABE = \alpha.$

Then $cos(90+\alpha) = cos90cos\alpha - sin90sin\alpha$ $= -sin\alpha$ $= \frac{5}{13}$

Finally, using law of cosines, we get:

$EF^2 = 130^2 + 238^2 + 2(130)(238)(\frac{5}{13}) = 312^2$

$EF = 312$

Observe that $\overline{AE}$ is the perpendicular bisector of $\overline{BC}$ , since $\overline{AB}=\overline{AE}=338.$ Then $\overline{EB}=\frac{1}{2}\cdot 260=130.$

By The Pythagorean Theorem, $\overline{AE}=\sqrt{338^2-130^2}=312.$

Now observe that quadrilateral $EAFC$ is cyclic with $\overline{AC}$ as diameter, since $\triangle{AEC}$ and $\triangle{AFC}$ are right triangles with $\overline{AC}$ as hypotenuse.

Then it follows that $\angle CAE=\angle EFC=\frac{1}{2}\stackrel\frown{EC.}$

Since $\angle BAE=\angle CAE$ by angle bisector $\overline{AE}$ , $\angle BAE=\angle EFC=\alpha.$

Given that $\angle BAF=\angle CFA=90$ ,

$\angle EAF =\angle BAF-\angle BAE=90-\alpha$ and $\angle EFA =\angle CFA-\angle EFC=90-\alpha.$

Then $\angle EAF= \angle EFA$

$\Rightarrow \triangle AEF$ is isosceles

$\Rightarrow \overline{EF}=\overline{AE}=\fbox{312}.$

We divide all lengths by the common factor 26 so that |AB| = |AC| = 13, |BC| = 10 Note that triangle CAB is isoceles, so angle AEB is right, and E is the midpoint of BC (making |BE| = 5).

Draw a line through E parallel to AB and CD, intersecting AF at M. Since E is the midpoint of BC, M is the midpoint of AF. Since EM is perpendicular to AF (which was perpendicular to CD), triangles AME and FME are right, and congruent since they share sides |ME| and |AM| = |FM|. Thus |EF| = |AE|, which is 12 by Pythagoras on right triangle AEB.

Scaling back up by 26 gives |EF| = 312.

ABC is isosceles triangle and AE is the height of ABC.

AE is the perpendicular bisector of BC, so CE = 130.

As per Pythagoras theorem,

AE^2 = AC^2 - CE^ = 338^2 - 130^2 = 312^2, so AE = 312

Now [ADC] = [ABC] , so

CD * AF = BC * AE , so 338 * AF = 260 * 312

hence AF = 240

Now for triangle AEF, AE = 312, AF = 240.

angle ACB = angle ABC = x. (Isosceles triangle)

angle ABC = angle ADC = x (opposite angles of parallelogram)

cosx = CE / AC ---> a)

angle ACD = y

Adjacent angle of parallelogram sums to 180, so

angle ADC + angle BCD = 180

x + angle BCA + angle ACD = 180

2x + y = 180 --> b)

Now,

angle DAF + angle FAE + angle EAB = x + y (opposite angles of parallelogram)

90-x + angle FAE + 90-x = x+y

angle FAE = 3x + y - 180 = x + 2x + 7 -180 = x ---> c)

For triangle FAE, using cosine rule

EF^2 = AF^2 + AE^2 - 2 * AF * AE * cos(angle FAE)

      = 240^2 + 312^2 - 2*240*312*cosx

      = 240^2 + 312^2 - 2*240*312* CE / AE

      = 240^2 + 312^2 - 2*240*312* 130 / 338

EF^2 = 312^2

so EF = 312

Since AB = AC, triangle ABC is isosceles with base BC. Thus, if E is the foot of the perpendicular from A to BC, E is the midpoint of BC, so CE = 260/2 = 130.

Similarly, CAD is isosceles. Let G be the foot the perpendicular from C to AD. Then, AG = 130 and AC = 338. Note that CGA is a 5-12-13 triangle, scaled up by a factor of 26, so CG = 12 * 26 = 312.

The area of triangle ACD is 1/2 * CG * AD = 1/2 * AF * CD, so:

AF = CG * AD / CD = 312 * 260 / 338 = 240

Since AFC is also a right triangle, and since AF = 240 and AC = 338, FC = 238 as the last leg of the triangle.

Consider quadrilateral AECF. Since both angles AEC and CFA are right, AECF is cyclic. Thus, we can use Ptolemy's Theorem:

AC * EF = AE * FC + CE * FA

338 * EF = 312 * 238 + 130 * 240 = 105456

EF = 105456 / 338

EF = 312

Since, by construction, $\angle AFC=\angle AEC=90^\circ$ , quadrilateral $AECF$ is cyclic.

Since $AC=AB$ , triangle $ABC$ is isosceles, so $E$ is the midpoint of $BC$ . Therefore, $BE=EC=130$ . From Pythagorean Theorem on $ABE$ , $AE=312$ .

Now, the area of triangle $ABC$ is $\frac{1}{2}\cdot AE\cdot BC=43940$ , and this is also equal to the area of triangle $ACD$ . Therefore, $\frac{1}{2}\cdot AF\cdot CD=43940$ , and so $AF=240$ . Now from Pythagorean Theorem on triangle $AFC$ , $FC=238$ .

Finally, since $AFCE$ is cyclic, we can use Ptolemy's, which states that $AF\cdot EC+AE\cdot FC=AC\cdot EF$ , from which we find that $EF=312$ .

    Since <AEC = <AFC = 90, AECF is a cyclic quadrilateral. It is given that AC = 338. Also, since ABC is A-isosceles with BC = 260, we find that EC = 130. We can solve AE using the Pythagorean Theorem to find that AE = 312. Finally, we can use the Pythagorean Theorem on A,C,F, and D to find that FC = 238 and AF = 240. 
    Now we can apply Ptolemy's Theorem on quadrilateral AECF. 
    AE*FC + AF*CE = AC*EF
    312*238 + 240*130 = 338*EF
    Solving gives EF = 312

First note that AECF is cyclic because <AEC and <CFA are both 90 degrees.

ABC is isosceles; this makes finding AE much easier as AE is both an altitutde and a median. EB = 130 = 26 5, AB = 338 = 26 13; so AE = 26*12 = 312.

Let CF = a, DF = b.

338^2 - a^2 = 260^2 - b^2 a^2 - b^2 = 338^2 - 260^2 a - b = 338 - (260^2/338) = 138 (since a+b = 338)

Therefore a = CF = 238, FA = sqrt(260^2 - b^2) = sqrt((360)(160)) = 240.

By ptolemy's theorem,

EF = ((CF AE) + (AF EC))/AC = (238 * 312 + 240 * 130)/338 = (119 * 156 * 4 + 50 * 156 * 4)/338 = 2 * 156 = 312.

I decided to place $ABCD$ on a coordinate plane because I saw that side $AD$ was $260$ , meaning that we can construct a $100:240:260$ right triangle (from the Pythagorean triple of $5:12:13$ ).

Thus, we have point $D$ at $(0,0)$ , point $A$ at $(100, 240)$ , point $B$ at $(438, 240)$ , and point $C$ at $(338, 0)$ . It's easy to see that point $E$ is at $(100, 0)$ . Point $F$ can be found by drawing a perpendicular line through $A$ to $CB$ , giving us $(388, 120)$ for $F$ .

Using distance formula, we then have that $EF = \sqrt{(388-100)^2 + (120 - 0)^2} = 312$ , which is our answer.

AE=sqrt(338^2-130^2)

Since AB=338=AC, triangle ABC is an isosceles triangle with angle ACE=angle ABE. E is the midpoint of BC and EC=BC/2=260/2=130. By Pythagoras' Theorem, AE^2=AC^2-EC^2=338^2-130^2=97344 and AE=312. Since ABCD is a parallelogram, angle ADF=angle ABE=angle ACE. Since angle ADF=angle ACE, angle AFD=90=angle AEC, triangle AFD is similar to triangle AEC. Consequently, DF/CE=AD/AC, DF/130=260/338, DF=100. FC=CD-DF=AB-DF=338-100=238. By Pythagoras' Theorem again, AF^2=AC^2-FC^2=338^2-238^2=57600 and AF=240. Since angle AFC+angle AEC=90+90=180, AECF is a cyclic quadrilateral. By Ptolemy Theorem, (EF)(AC)=(AF)(EC)+(AE)(FC), (EF)(338)=(240)(130)+(312)(238), EF=312.

in IIgm ABCD, (by geometry) by trigonometry AE = 312 now,- let angle ABC be "x". using trigo we will get AF in terms of x. on solving further and using sum of interior angles of a triangle is 180, we will get angle EAF also "x". now, using simple trigonometry we will get cosx=5/13 and sinx = 12/13 now, by using cosine formula in triangle EAF and substituting the value of sides and cosx we will get EF=312

Since ABCD is a parallelogram, we have BA \ CD and BC \ AD. Thus <BAC = <ACD and <BCA = <CAD. Since ABC is isosceles, so is ACD. Therefore CD = AB = 338, BC = AD = 260.

Since ABC is isosceles, E is the midpoint of C. Thus BE = EC = 130. By Pythagoras' Theorem on triangle ABE, AE = sqrt(338^2 - 130^2) = 312. Now Area of ABC = Area of ACD, so (1/2)(312)(260) = (1/2)(Length of AF)(338). Solving, we have AF = 240.

Using Pythagoras Theorem on AFD, we have FD = sqrt(260^2 - 240^2) = 100. Now CF = 338 - 100 = 238.

Since <AEC + <AFC = 180 degrees, we have AECF is a cyclic parallelogram.

Now using Ptolemy's Theorem on AECF, we have (338)(Length of EF) = (130)(240) + (312)(238)

Solving, we have EF = 312.

HERE AB IS 338 BC IS 260 AC IS 338 AD IS 260 DC IS 338 SO AE IS PERPENDICULAR BISECTOR SO EC =130 BY PHYTHAGOS IN ABE AE =312 SO AREA = AE BC=312 260 ALSO AF*DC=AREA SO AF =240 BY PHYTHAGORAS THEOREM IN AED !WE GET DF=100 SO FC=238 NOW USING FC=238 AND EC=130 IN EFC WE GET EF =312

Observe that $338 = 26 \cdot 13$ and $260 = 26 \cdot 10$ . For ease of calculation let's shrink $ABCD$ by a factor of $26$ . Since $BC = AC = AD$ , $\triangle DAC$ is isosceles, so $AF$ is both an altitude and median. $FC = 5$ and by the Pythagorean theorem $AD = 12$ . Opposite angles of parallelograms are congruent, so $\angle ADC = \angle ABC$ . By angle-angle $\triangle AEB$ is similar to $\triangle AFC$ , so $AE = \frac {10}{13} \cdot 12 = \frac {120}{13}$ , and $CE = 13 - \frac {10}{13} \cdot 5 = \frac {119}{13}$ . Being at intersections of perpendicular lines, $\angle AFC = \angle AEC = 90 ^ \circ$ , so quadrilateral $AECF$ is cyclic. By Ptolemy's theorem $12 \cdot \frac {119}{13} + 5 \cdot \frac {120}{13} = 13 \cdot EF$ . Solving we get $EF = 12$ . Dilate by the factor of $26$ to get $EF = 312$ .

Since ABC is an Isosceles Triangle with sides AB = AC , $\angle ABC$ = $\angle ACB$ = $\cos ^ {-1} \frac {(260/2)}{338}$ = $67.38 ^ \circ$ . Now, $\angle ACF$ = $\angle BAC$ = $45.24 ^ \circ$ . We now find the sides of triangle AEF . Length AE = $AB\sin 67.38 ^ \circ$ = 312. Length AF = $AC\sin 45.24 ^ \circ$ = 240. $\angle EAF$ = $\angle EAC + \angle CAF$ = $\frac {45.24}{2} + (180 ^ \circ - \angle ACF)$ = $67.38 ^ \circ$

Therefore, by applying COSINE RULE , we get $EF ^ 2$ = $AE ^2 + AF ^2 - 2 \times AE \times AF \times cos \angle EAF$ = 97344

EF = $\sqrt{97344}$ = 312

We will show that $AEF$ is similar to $BAC$ . Let $S$ be the area of triangle $ABC$ . Since $ABC$ and $CDA$ are congruent triangles, $S=[ABC]=[ACD]$ . Using the formula for area, we can calculate that $AE = \frac {2S}{BC}$ and $AF = \frac {2S} {CD}$ , hence $\frac {EA}{AF} = \frac {CD}{BC} = \frac {AB}{BC}$ . From quadrilateral $AECF$ , $\angle EAF = 360^\circ - \angle AEC - \angle ECF - \angle CFA$ $= 360^\circ - 90^\circ - 90^\circ - \angle ECF$ $= 180^\circ - \angle ECF$ $= 180^\circ - \angle BCD$ $= \angle ABC$ . Thus, the triangles $AEF$ and $BAC$ have sides that are proportional to each other, and an equal angle between these proportionate sides, hence are similar by side-angle-side.

Since $ABC$ is isosceles and $AB=AC$ , thus $EF=EA$ by similar triangles. Since $AE$ is the height of isosceles triangle $ABC$ , then $CE = \frac{BE}{2} = 130$ . So, we can calculate that $AE^2 = AC^2 - CE^2 = 338 ^2 - 130 ^2 = 26 ^2 (13^2-5^2) =26 ^2 \cdot 12^2$ $\Rightarrow AE = 26\cdot 12 = 312$ . Thus, $EF = AE = 312$ .

Note: $AEF$ is similar to $BAC$ regardless of the dimensions of the parallelogram.