Perpendicular Lines in Triangle

Connect points $D$ and $E$ . Triangles $DEC$ and $BAC$ are similar by $SAS$ , so $DE$ is parallel to $BA$ . Let the intersection of $AD$ and $BE$ be $F$ . Triangles $DFE$ and $AFB$ are similar by $AA$ and the similarity ratio is $\frac {1} {2}$ because $DE$ is $\frac {1} {2}$ of $BA$ . By the Pythagorean theorem, $DF^2+(2FE)^2=89^2$ , $FE^2+(2DF)^2=69.5^2$ , and $DF^2+ FE^2= DE^2$ . Adding the first two equations and dividing by $5$ gives $DF^2+FE^2=DE^2=2550.25$ . So $DE=50.5$ and $BA=101$ .

This Wikipedia entry on Median's Equal-area division contains a proof that the 3 medians of $\triangle ABC$ will divide the triangle into 6 smaller triangles of equal area. In particular for us, we are interested in the result that $2[FBD] = 2[FAE] = [FAB]$ .

Since $\angle AFB = 90^\circ$ , the usual area formula for triangle give us $2 FB \cdot FD = 2 FA \cdot FE = FA \cdot FB$ , which simplifies to $2 FD = FA$ and $2FE = FB$ .

Pythagorean theorem seals the deal for us with $AC^2 = 4 AE^2 = 4 FA^2 + 4 FE^2 = 4 FA^2 + FB^2$ ; $BC^2 = 4 BD^2 = 4 FD^2 + 4 FB^2 = FA^2 + 4 FB^2$ , and so $AC^2 + BC^2 = 5 \left( FA^2 + FB^2 \right) = 5 AB^2$ .

Lastly, we substitute in the values to get $AC^2 = \sqrt{\dfrac{139^2 + 178^2}{5}} = 101$ .

Draw the picture. Assume that, $BE$ and $DF$ intersect at $F$ . Also assume that, $EF = x$ , and $FD = y$ .

Also, $AE = \frac{AC}{2} = \frac{139}{2}$ and $BD = \frac{BC}{2} = 89$ .

As we know that, by median formula. We obtain that :

$$ $BF = 2(EF) = 2x$ and $AF = 2(FD) = 2y$ $$ .

From the right $\triangle BFD$ and $\triangle AFE$ , we obtain :

$x^2 + 4y^2 = \frac{139^2}{4} \ldots \ldots (1)$

$y^2 + 4x^2 = 89^2 \ldots \ldots (2)$

Then, sum $(1)$ and $(2)$ , we obtain :

$5(x^2 + y^2) = \frac{51005}{4} \implies x^2 + y^2 = \frac{10201}{4}$ .

Now, we see that the right $\triangle AFB$ , then :

$AF^2 + BF^2 = AB^2$ $4x^2 + 4y^2 = AB^2$ $4(\frac{10201}{4}) = AB^2 \implies AB^2 = 10201 \implies AB = \boxed{101}$

First, we connect the points E and D . We let F be the intersection of AD and BE .

Given that AD and BE are perpendicular to each other, we will have the triangles AFE , BFD , AFB and EFD which are all right at F .

Since D and E are the midpoints of BC and AC respectively, it follows that DB and EA are equal to 89 and $\frac {139}{2}$ respectively.

Considering the right triangles AFE and BFD , and applying the Pythagorean Theorem, we will have the following equations;

(1) $AF^2$ + $EF^2$ = $\frac {19321}{4}$

(2) $BF^2$ + $DF^2$ = 7921

Adding the equations (1) and (2), we will have;

(3) $AF^2$ + $BF^2$ + $EF^2$ + $DF^2$ = $\frac {51005}{4}$

Now, considering the right triangles AFB and EFD , and applying the Pythagorean Theorem, we will have the following equations;

(4) $AF^2$ + $BF^2$ = $AB^2$

(5) $EF^2$ + $DF^2$ = $ED^2$

Substituting (4) and (5) to (3), we will have the equation;

(6) $AB^2$ + $ED^2$ = $\frac {51005}{4}$

Given that D and E are the midpoints of BC and AC respectively, and applying the Mid-Segment Theorem, we can say that ED is $\frac {AB}{2}$ .

Replacing ED by $\frac {AB}{2}$ in (6), we will have the equation;

(7) $AB^2$ + $\frac {AB^2}{4}$ = $\frac {51005}{4}$

Multiplying both sides of the equation by 4, we will have;

(8) $4AB^2$ + $AB^2$ = 51005

By addition, we will have;

(9) $5AB^2$ = 51005

Dividing both sides by 5, we will have;

(10) $AB^2$ = 10201

We extract the roots of both sides of the equations and considering only the positive solution (since measurements of length are positive) we will have;

AB = 101

The length of AB is 101.

Let the centroid be F. Note that F is the intersection of AD and BE.

We connect DE. Since CE/CA=CD/CB, and <ECD=<ACB, ECD and ACB are similar. Thus ED || AB and ED:AB=1:2. Note that <DEF=<FBA and <EDF=<FAB. Thus by AA similarity, EDF and BAF are similar.

Since ED:AB=1:2, EF/FB=1/2 and DF/FA=1/2. We let EF=x and DF=y. Thus FB=2x and FA=2y.

By the Pythagorean Theorem, BD^2=DF^2+FB^2=4x^2+y^2. BD^2=(178/2)^2=7921. AE^2=EF^2+FA^2=x^2+4y^2. EF^2=(139/2)^2=4380.25.

Thus we have the following equations: 4x^2+y^2=7921 x^2+4y^2=4380.25

Note that AB^2=(2x)^2+(2y)^2=4x^2+4y^2. Adding up the 2 equations we get 5x^2+5y^2=12751.25. Hence AB^2=4x^2+4y^2=(12751.25)/5*4=10201.

Hence AB=sqrt(10201)=101.

Since the two medians intersect at right angles, say at point $F$ within the triangle, $(1)$ three right triangles are formed: $\triangle BFD$ , $\triangle EFA$ , and $\triangle AFB$ ; and $(2)$ $F$ is two-thirds of the way from every vertex of $\triangle ABC$ . We can let $x = FE$ and $y = FD$ , so that $BF = 2x$ and $AF = 2y$ .

Using the Pythagorean Theorem on $\triangle BFD$ , we have

$(A) 4x^2 + y^2 = 89^2$

From $\triangle EFA$ , we have

$(B) x^2 + 4y^ 2 = (\frac {139} {2})^2$

Also, from $\triangle AFB$ , we have

$(C) 4x^2 + 4y^ 2 = AB^2$

We don't need the exact values of $x$ and $y$ for this problem, since we are only asked to find the value of $AB$ , which is equal to $\sqrt {4x^2 + 4y^2}$ . So it suffices to find the values of $y^2$ and $x^2$ .

Plugging in the expression for $x^2$ in terms of $y^2$ from $(B)$ to $(A) (x^2 = (\frac {139} {2})^2 - 4y^2)$ , we can get $y^2 = 760$ , and $4y^2 = 3040$ . Replacing $y^2$ by $760$ in $(A)$ , we get $4x^2 = 7161$ . Thus, from $(C)$ , $AB = \sqrt {3040 + 7161} = \sqrt{10201}$ , or $\boxed {AB = 101}$ .

Since E and D are midpoints of side AC and BD, therefore $2ED=AB$ .

Let G be the intersection of line AD and BE.

By pythagoras theorem, we have:

$GD^2+GE^2=ED^2$

$GE^2+GA^2=(\frac{139}{2})^2=4830.25$

$GA^2+GB^2=AB^2=4ED^2$

$GB^2+GD^2=(\frac{178}{2})^2=7921$

Combining first and third equation, we have the sum of squares of all sides is $5ED^2$ . Combining second and fourth equation, we have the sum of squares of all sides is 12751.25. Therefore, $5ED^2=12751.25$ . So, $ED=50.5$ and $AB=101$ .

The property of a line joined by 2 mid points of a triangle's sides is that it will be parallel to the third line.

In this problem we are given that $D$ and $E$ are midpoints of $BC$ and $AC$ respectively, which according to the property, tell us that $DE$ is parallel to $AB$

Also, let the intersection between $BE$ and $AD$ be $F$

According to similarity $\frac{DE}{BA}= \frac{CD}{CD} = \frac {CE}{CA}$ which gives us the value $\frac{DE}{BA} = \frac{1}{2}$

Consider the triangles $ABF$ and $DEF$ . $AB$ is parallel to $DE$ so $\angle FBA = \angle FED$ and $\angle FAB = \angle FDE$ and we know that $\angle AFB = \angle DFE = 90^\circ$ so $ABF$ is similar to $DEF$

so $\frac{DE}{AB} = \frac {EF}{BF} = \frac{FD}{FA}$

Let $FE = a$ and $FD = b$ , so $BF = 2a$ and $AF = 2b$

according to Pythagorean theorem, the following equations hold :

$AF^2 + FE^2 = AE^2 \rightarrow (2b)^2 + (a)^2 = 69.5^2$ ...(1)

$BF^2 + FD^2 = BD^2 \rightarrow (2a)^2 + b^2 = 89^2$ ...(2)

$BF^2 + AF^2 = AB^2 \rightarrow (2a)^2 + (2b)^2 = AB^2$ ...(3)

(1)+(2)

$5a^2 + 5b^2 = 12 751.25$ .. multiply both sides by $0.8$

$4a^2 + 4b^2 = 10 201$

Subtituting (3)

$AB^2 = 10 201 \rightarrow AB= 101$

So the value of $AB$ is $101$

Let $G$ be the intersection of $AD$ with $BE$ .

Since $D$ and $E$ are midpoints of $BC$ and $AC$ respectively, segments $AD$ and $BE$ are medians of the triangle, which in turn makes $G$ be the centroid of triangle $ABC$ .

For the sake of simplicity, let $GD = a$ and $GE = b$ .

Now, it is known that the centroid divides each median into parts in the ratio 2:1, which means $AG = 2 GD = 2a$ and $BG= 2GE=2b$ .

Since $AD$ and $BE$ are perpendicular to each other, we can use the Pythagorean Theorem on triangles $ABG$ , $BGD$ and $AGE$ . We get from triangle $ABG$ that $AB^2 = BG^2+AG^2=4b^2+4a^2 .$

By the Pythagorean theorem on triangles $BGD$ and $AGE$ we obtain $BD^2=BG^2+GD^2=4b^2+a^2$ $AE^2=AG^2+GE^2=4a^2+b^2$ respectively. It's clear that $BD=\frac{BC}{2}=89$ and that $AE=\frac{AC}{2}=69.5$ . In other words, $89^2 = 4b^2+a^2$ and $69.5^2 = 4a^2+b^2.$ Adding these equations we get $69.5^2+89^2=5(a^2+b^2).$ Multiplying this last equation by $\frac{4}{5}$ leads to

$AB^2=4(a^2+b^2)= (69.5^2+89^2)(\frac{4}{5}) =10201.$ Therefore, $AB=101$ .

LET AD AND BE INTERSECT AT O. SINCE AD AND BE ARE MEDIANS THEREFORE POINT O MUST BE THE CENTROID . AS WE KNOW THAT CENTROID DIVIDES THE MEDIAN IN THE

RATIO OF \frac {2}{1} , THEREFORE , LET AO=2X, OD =X BO=2Y,OE=Y

SINCE AD AND BE ARE PERPENDICULAR TO EACH OTHER THEREFORE,

AO^2+OE^2=AE^2 AND BO^2+OD^2=BD^2

SO, 4X^2+Y^2=(69.5)^2 AND 4Y^2+X^2=89^2

ADDING THE TWOEQUATIONS , WE GET 5(X^2+Y^2)=\frac {51005}{4} SO, 4(X^2+Y^2)=\frac {51005}{5} SO,AB^2=10201 AND HENCE AB =101.

Let $AD$ and $BE$ intersect at $G$ , which is the centroid of the triangle. We know that $AG = 2 GD$ and $BG = 2 GE$ .

Applying the Pythagorean theorem on triangles $BGD$ and $AGE$ , we get that

$4 GD^2 + GE^2 = \left( \frac{AC}{2} \right) ^2,\ \ 4 GE^2 + GD^2 = \left( \frac{BC}{2} \right) ^2.$

Hence, $AB^2 = 4 GD^2 + 4 GE^2 = \frac{ 4}{5} \left[ \left( \frac{AC}{2} \right) ^2 + \left( \frac{BC}{2} \right) ^2 \right] = 10201$ . Thus, $AB = 101$ .

. Let F is the intersection point of AD and BE. Triangle EDC is similar to ABC (length ratio 1:2). Let EF=a, and FD=b. a2 + 4b2 = (139/2)2, b2 + 4a2 = (178/2)2, So a2 = 1790.25, b2 = 760, AB2 = 4a2 + 4b2 = 10201, AB = 101.

The formula for this question is square root (a^2+b^2)/2=squreroot (10201)=101.

Let the point where $AD$ and $BE$ intersect be F.

Let the length $AF$ be $x$ and $BF$ be $y$ .

Since each of the medians divide one another in the ratio 1:2, we can write $DF$ as $\frac{x}{2}$ and $EF$ as $\frac{y}{2}$ .

By Pythagoras' Theorem, we have, after simplifying:

$4x^{2}+y^{2}=139^{2}$ $$ $$ $$ (Equation 1 from triangle $AFE$ )

$x^{2}+4y^{2}=178^{2}$ $$ $$ $$ (Equation 2 from triangle $BFD$ )

Adding the two equations together,

$5x^{2}+5y^{2}=51005$

$x^{2}+y^{2}=10201$

Since $AB$ , the required length, is the hypotenuse of right-angled triangle $ABF$ with edges of length $AF=x$ and $BF=y$ , Pythagoras' Theorem gives us the final answer:

$\sqrt{x^{2}+y^{2}}=\boxed{101}$

I used a coordinate system with the intersection of BE and AD as the origin, and E, D on the x-axis and y-axis, respectively. The coordinates of the various points may now be written as $A\ (0,-a);\ B\ (-b, 0);\ D\ (0, d);\ E\ (e, 0),$ and from the fact that D and E are midpoints, $C\ (b, 2d) = (2e, a),$ so that $d = \tfrac a2, e = \frac b2$ . The given lengths show that $a^2 + e^2 = \tfrac14\cdot 139^2,\ b^2 + d^2 = \tfrac14\cdot 178^2;$ we must find the square root of $a^2 + b^2 = (a^2 + e^2 + b^2 + d^2) - (b^2 + d^2) = \tfrac14(139^2 + 178^2) - \tfrac14(a^2 + b^2),$ $a^2 + b^2 = \frac{139^2 + 178^2}5 = 10201,$ from which we find $AB = \sqrt{a^2 + b^2} = \boxed{101}.$