In triangle $ABC$, $BC = 4$, $AC = 3 \sqrt{2}$, and $\angle C = 45^\circ$. Altitudes $AD$, $BE$, and $CF$ intersect at the orthocenter $H$. Find $AH:HD$.
I tried using the pythagorean theorem to find side lengths, algebraically, but I couldn't find anything. Help!
Yes, I agree!
I think this will help you out:
http://jwilson.coe.uga.edu/EMAT6680Fa2012/Szatkowski/SzatkowskiWU8/ASwriteup8.html
As in all geometry problems,
1. start with drawing a diagram indicating all given information.
2. deduce from known theorems additional information, and mark any new information on the diagram.
3. repeat #2 above.
I have done the first step for you, and the diagram is located at:
https://www.twiddla.com/fb5es1
Hints:
Note that congruent angles have been marked in read (step 2).
study the right-triangle BCE.
hence calculate mCE, then mEA (mark on diagram).
Study the right-triangle EAH, and deduce length of mAH (mark on diagram).
Similarly, study the right-triangle CAD.
Calculate mCD, hence mDB (mark on diagram).
Study right-triangle DBH, deduce length of mHD (mark on diagram).
Finally calculate ratio mAH:mHD.
stop cheating on Aops alcumus
To find the ratio $AH:HD$, we can start by finding the lengths of $AH$ and $HD$ individually.
Let's first find the length of $AH$. Since $AH$ is an altitude, it is perpendicular to the base $BC$. In triangle $ABC$, we have a right triangle formed by $AH$, $BC$, and $AB$.
Using the given information, we can find the length of $AB$ using the Pythagorean theorem. The side of length $AC$ is the hypotenuse of this right triangle, and we know that $AC = 3\sqrt{2}$.
Applying the Pythagorean theorem, we have:
$AC^2 = AH^2 + CH^2$
$(3\sqrt{2})^2 = AH^2 + (BC)^2$
$18 = AH^2 + 16$
$AH^2 = 2$
Taking the square root of both sides, we get $AH = \sqrt{2}$.
Now, let's find the length of $HD$. Since $HD$ is also an altitude, it is perpendicular to the base $AC$. In triangle $ADC$, we have a right triangle formed by $HD$, $AC$, and $AD$.
Using the given information, we can find the length of $AD$ using the Pythagorean theorem. The side of length $BC$ is the hypotenuse of this right triangle, and we know that $BC = 4$.
Applying the Pythagorean theorem, we have:
$BC^2 = HD^2 + CD^2$
$4^2 = HD^2 + (AC)^2$
$16 = HD^2 + (3\sqrt{2})^2$
$16 = HD^2 + 18$
$HD^2 = -2$
Uh-oh! It seems there might have been an error in the calculation. Let's double-check our work.
Upon closer inspection, we realize that we made a mistake when applying the Pythagorean theorem in triangle $ADC$. We accidentally squared the lengths of the sides $AC$ and $BC$, which resulted in an incorrect calculation.
Instead, we should have applied the Pythagorean theorem correctly:
$AD^2 = HD^2 + CD^2$
Since we don't know the length of $AD$, we need to find it using the given information. To do this, we can use another right triangle: $ABC$.
In triangle $ABC$, $\angle C = 45^\circ$. Thus, triangle $ABC$ is a 45-45-90 triangle. In a 45-45-90 triangle, the sides are in the ratio $1:1:\sqrt{2}$. Since the hypotenuse $AC$ is given as $3\sqrt{2}$, the other sides $AB$ and $BC$ must also be $3\sqrt{2}$ each.
Now we can use the Pythagorean theorem in triangle $ADC$:
$AD^2 = HD^2 + CD^2$
$(3\sqrt{2})^2 = HD^2 + (4)^2$
$18 = HD^2 + 16$
$HD^2 = 2$
Taking the square root of both sides, we get $HD = \sqrt{2}$.
Now that we have calculated the lengths of $AH$ and $HD$, we can find the ratio $AH:HD$:
$AH:HD = \sqrt{2}:\sqrt{2} = \boxed{1:1}$.