**Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove:**

**HD/AD + HE/BE + HF/CF = 1 and AH/AD + BH/BE + CH/CF = 2.**

To prove these relationships, we will use area of triangles and ratios. First we will prove the relation equating to 1.

Proof:

We first note that triangle ABC must be acute; that is, any angle of the triangle must be less than or equal to 90°.

For triangle ABC, we know the area A = 1/2*BC*AD = 1/2*AC*BE = 1/2*AB*CF, where AD, BE, and CF are the altitudes to their respective bases. We will let A_{1} be the area of triangle BHC, A_{2} be the area of triangle AHB, and A_{3} be the area of triangle AHC, where H is the orthocenter of triangle ABC. Then, A_{1} = 1/2*HD*BC, A_{2} = 1/2*HF*AB, and A_{3} = 1/2*HE*AC. We know that the sum of the areas of these triangles is equal to the area of triangle ABC; thus, A = A_{1} + A_{2} + A_{3}. If we divide through both sides with A, the area of the original triangle, then we have 1 = A_{1}/A + A_{2}/A + A_{3}/A. Now we coordinate values of A based on the values of each area of the smaller triangles. Thus, A_{1}/A = (1/2*HD*BC)/(1/2*BC*AD) = HD/AD. Next, A_{2}/A = (1/2*HF*AB)/(1/2*AB*CF) = HF/CF. Lastly, A_{3}/A = (1/2*HE*AC)/(1/2*AC*BE) = HE/BE. Thus, we have that 1 = HD/AD + HF/CF + HE/BE.

Now we much show that 2 = AH/AD + BH/BE + CH/CF. We already have that 1 = HD/AD + HF/CF + HE/BE. Note that HD = AD - HA, HF = FC - HC, and HE = BC - HB. Then we use these in substitution with the previously found relation. Therefore, 1 = (AD - HA)/AD + (FC - HC)/FC + (BE - HB)/BE = 1 - HA/AD + 1 - HC/FC + 1 - HB/BE = 3 - (HA/AD + HC/FC + HB/BE). Now we have that 1 = 3 - (HA/AD + HC/FC + HB/BE); hence, 2 = HA/AD + HC/FC + HB/BE.

QED