Triangle from Perpindiculars

Consider the nine-point circle of triangle $ABC$ . From symmetry arguments, this circle is tangent to $BC$ at $D$ , and since the nine-point circle passes through the midpoint $J$ of $AH$ , $JD$ is a diameter. Since $E$ , the foot of the altitude from $B$ to $AC$ , also lies on the circle, $\angle JED=90^\circ$ .

Since $AB=AC$ , then the perpendicular line from $A$ to $BC$ , bisects $BC$ , therefore $D$ is midpoint of $BC$ . Since triangle $BEC$ has a right angle in $E$ , and $D$ is midpoint, then $DB=DE=DC$ , so triangle $BDE$ is isosceles, $\angle DEB= \angle DBE$ . Since $AHE$ , has a right angle at $E$ , and $J$ is midpoint, then $JE=JH=JA$ , then triangle $JHE$ is isosceles, therefore $\angle JEH= \angle JHE= \angle BHD$ , so $\angle JED=\angle JEH+\angle HED= \angle BHD+ \angle HBD= 90^\circ$ .

Notice that, by the nine-point circle theorem, points $J,E,D$ lie on the nine-point circle with centre $K$ , the midpoint of $OH$ , where $O$ is the circumcentre of $\bigtriangleup ABC$ . Since $AB=AC$ , points $O,H$ lie on line $AD$ . So $JD$ is diameter of circle. Since $E$ lies on the nine-point circle, $\angle JED=\frac{180^{\circ}}{2}=90^{\circ}$ .

The Nine Point Circle of this triangle passes through J, E, and D. By symmetry, JD is the diameter of this circle (or rather, we can let F be the midpoint of AB and AD is the perpendicular bisector of EF, hence implying JD passes through the center of the circle). Therefore angle JED is 90 degrees.

Consider the nine point circle of triangle $ABC$ . Then note that the center of it lies on line $AD$ because $O$ lies on $AD$ and the nine-point center is the midpoint of $HO$ . But then it follows $JD$ is a diameter of the circle, and as $E$ lies on the nine point circle we have $\angle JED = 90$ as desired.

$J$ is the midpoint of $AH$ . Since $E$ is the foot of the perpendicular from $B$ , triangle $AEH$ is a right triangle, with $EJ$ being the median of the hypotenuse. Therefore, $EJ$ is half the length of $AH$ , or $EJ = AJ$ . This makes triangle $AJE$ isosceles, and $\angle AEJ = \angle JAE$ .

Since $AB=AC$ , and $D$ is the foot of the perpendicular from $A$ , $BD=DC$ . Triangle $BEC$ is right, with $ED$ being the median to the hypotenuse. Therefore $ED=DC$ , and triangle $EDC$ is isosceles. Thus $\angle DEC = \angle DCE$ .

Furthermore, triangle ADC is right, so $\angle JAE + \angle DCE = 90^\circ$ . Therefore $\angle AEJ + \angle CED = 90^\circ$ . This leaves $\angle JED = 90^\circ$ .

First, note that $\angle AHE = 180^\circ-\angle DHE = \angle DCE = \alpha$ . Next, since $\angle HEA = 90^\circ$ , therefore $J$ is the circumcenter of triangle $HEA$ , therefore $|JH|=|JE|$ , thus $\angle JEH = \angle JHE = \alpha$ .

Next, since $\angle BDA = \angle BEA = 90^\circ$ , $BDEA$ is a cyclic quadilateral, hence $\angle BED = \angle BAD = 90^\circ-\angle ABC = 90^\circ - \alpha$ , the last one since $ABC$ is isoceles.

So, we have $\angle JED = \angle JEH + \angle BED = \alpha + (90^\circ-\alpha) = 90^\circ$

Let's set angle EBC = x. From complimentary angles, angle BHD = 90 - x = x', since triangle BHD is right. Angle JHE is also x' because it is an opposite angle. Now, we observe that JE is a median to right triangle AHE, and therefore segments JE = JH because a median to the hypotenuse of a right triangle is equal to half the hypotenuse (this can be seen from the circumscribed circle: the hypotenuse is the diameter and the median is a radius). Therefore, angle JEB is also x' as triangle JEH is isosceles. Repeating this procedure with right triangle BEC, segments ED = BD as ED is a median to the hypotenuse. This is because the altitude AD is also a median of isosceles triangle ABC. From here, angle BED = x because triangle BED is isosceles. Therefore, angle JED = angle JEB + angle BED = x + x' = 90 degrees.

First, we note the following facts (which can be easily proved.) :

i) A triangle is isosceles if and only if its altitude to its incongruent side is the angle bisector and median.

ii) Let $\triangle ABC$ be a right triangle with right angle at B and let BD be the median of $\triangle ABC$ , then $BD = AD = CD$ .

iii) Let $\triangle ABC$ be an isosceles triangle, and let $AD, BE, CF$ be its altitudes. Prove that $\angle FEB = \angle EBD$

We will not show i) and ii) since they are well-known facts, so let's show iii)

Let $\angle FEB = a$ . Let H be the orthocenter. Then $\triangle BHC$ and $\triangle FEH$ are isosceles. (This follows from i) and a bit of angle chasing.) Consequently, $\angle EFC = a, \angle FHE = 180 - 2a, \angle BHC = 180 - 2a$ , and from i), $\angle BHD = \angle DHC = 90 - a, \angle EBD = a$ .

Let CF be the third altitude. Let $\angle HCB = a$ , then from i) and angle chasing $\angle HBD = a, \angle BHD = 90 - a$ $\angle AHE = 90 - a, \angle HAE = a, \angle DAB = a$ . From ii), $\angle JEA = a, \angle JEH = 90 - a$ .

Now if you join AD, BE, CF to form $\triangle DEF$ , then it's a well-known fact that AD, BE, CF are the angle bisectors of $\triangle DEF$ . Let $\angle HED = b$ , then $\angle FEH = b$ and from iii) $\angle EBD = a = b$ , so $a = b$ .

Finally, $\angle JED = \angle JEH + \angle HED = 90^{\circ} - a + a = \boxed{90^{\circ}}$ .

The problem implies that for any isosceles triangle ABC, $\angle$ JED would remain constant. We may stretch its geometry so that BC = AB = AC, hence forming an equilateral triangle. The orthocenter H would now also be the centroid.

Since H is the centroid and J is the midpoint of AH, EH = HD = $\frac {1}{2}$ AH = JH = AJ.

On rt. $\Delta$ AEH, EJ is the median to the hypotenuse AH, thus, EJ = AJ.

By S.A.S, $\Delta$ ECD is also equilateral. Thus, $\Delta$ AEH $\cong$ $\Delta$ DEJ by S.S.S, making $\angle$ JED = 90 $^\circ$ .

BE is the altitude as well as the angle bisector of angle B. Therefore, angle HBD= angle HBA= theta. Angle BHD =angle AHE= alpha ( vertically opposite angles). From triangle HBD we get alpha +theta =90. Using the above relation we get angle EAH also equal to theta. Now, angle AHB+ angle AHE = 180( angle in a straight line). angle AHB =2 theta. We therefore get 2 theta+alpha=180 and by previous relation as alpha +theta =90, we get theta=30 and alpha =60.

In triangle JEH, on applying sine rule we get sin(JEH)/JH= sin(JHE)/JE. angle JHE = alpha=60 In triangle AJE, sin(AEJ)/JA= sin(JAE)/JE. angle AEJ= 90-JEH and angle JAE=theta=30. Also, JA=JH.On equating JE/JA from both the equations we get angle JEH=60.

Since,theta=30, we get the triangle as equilateral triangle. Therefore, HE=HD. Now, in triangle HED, angle DHE= 2 alpha=120. Therefore, angle HED=30. Hence,angle JED=angle JEH+angle HED=60+30=90.

By considering a triangle ABC with variables we can find this. Since AB=AC,angleABC=angleACB='x'. So angleBAC=180-2x. In the quadrilateral HEDC,two angles are 90 degrees and one angle=x. Therefore angleEHD=180-x. thus angleBHD in triangle HBD=x degrees (By Linear Pair). Thus by Angle Sum Property in triangle BHD,it can be found that x=45 degrees. By using x=45 degrees,we can find out other angles mentioned above. By finding these angles,we can frame a relation of these angles with angleJED and can find that angleJED=90 degrees.

J, D, E are points on the nine point circle of the triangle and JD is its diameter. Thus JED=90