IOQM/INMO- TRIANGLE GEOMETRY

Question

In triangle $ABC$ the altitude from $A$ to $BC$ meets $BC$ at $D$ and the altitude from $B$ to $AC$ meets $AD$ at $H.$ It is also given that $AD=4, BD=3, CD=2.$ Find $HD.$

This is an IOQM-INMO question and it is complete, though if you search on net it is also given that $AB/BD=AH/HD$ but I want to know if it can be solved without using this.

Answer 1

It's actually given in the question that $\frac{AB}{BD}=\frac{AH}{HD}$!! Without it it will be actually extremely difficult to solve it!

Anyway solving the problem-

$AB=\sqrt{AD^2+BD^2}=\sqrt{4^2+3^2}=5 cm$(According to Pythagorean theorem)

Now, Let $HD=x$ $\Rightarrow AH=4-x$

ATQ $\frac{AB}{BD}=\frac{AH}{HD}$ $\Rightarrow\frac{5}{3}=\frac{4-x}{x}$ $\Rightarrow 5x=12-3x$ $\Rightarrow x=\frac{3}{2}=1.5=HD$

Simple!

Answer 2

Yes, this problem can be solved without much ado and without resorting to use the additional information given in various websites. In $\mathrm{Fig.\space 1}$, all given segment values are marked in $\mathbf{\color{red}{red}}$. Denote the point of intersection between the altitude from the vertex $B$ and the side $AC$ as $E$.

By applying the theorem Pythagoras to the right angle triangle $ABD$, we obtain, $$AB=\sqrt{BD^2+DA^2}=\sqrt{3^2+4^2}=5. $$

Since the two sides $AB$ and $BC$ are equal in length, the triangle $ABC$ is an isosceles triangle. As a consequence the point $E$ is the midpoint of $CA$.

We apply Pythagoras’ theorem to the right angle triangle $ADC$, to obtain, $$AC=\sqrt{AD^2+DC^2}=\sqrt{4^2+2^2}=2\sqrt{5}.$$ Therefore, $CE=\sqrt{5}$.

Now, we apply Pythagoras’ theorem to the right angle triangle $BCE$, to get, $$EB=\sqrt{AC^2-CE^2}=\sqrt{5^2+\sqrt{5}^2}=2\sqrt{5}.$$

Finally, we turn our attention to the two right angle triangles $CEB$ and $BDH$. Since they share a vertex angle, namely $\measuredangle EBC$, they are similar, and hence we shall write, $$\frac{CE}{HD}=\frac{EB}{BD}\quad\rightarrow\quad \frac{\sqrt{5}}{HD}=\frac{2\sqrt{5}}{3}\quad\rightarrow\quad HD=\frac{3\times\sqrt{5}}{2\sqrt{5}}=1.5.$$

Here are some hints, which you can follow, if you encounter this type of problems. First, check whether, the given geometrical configuration is constructible using the data given in the problem. If so, then draw the figure to scale. Such a diagram can give you a wealth of information on how to proceed with to find a solution quickly.

Answer 3

After four weeks of cooldown time, I hope giving this answer will not raise any issue. So, as promised, here are three solutions, and you will find no square roots (except the 3-4-5 Pythagorean Triple).

I did not mark $AD = 4$ since the placement would be awkward.

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Method 1: The shortest solution: use the Angle Bisector Theorem.

Using $AC \perp BE$, $AB = BC$ and the common side $BE=BE$, $\triangle ABE \cong \triangle CBE$ by RHS. This gives $\angle ABE = \angle CBE$. Applying the theorem on $\triangle ABD$, we have $AB/BD = AH=HD$, which was the "not given" ratio.

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Method 2: The simplest solution: use similar triangles.

$\angle ADB = \angle ADC = 90^\circ$. We also have $\angle DBH = 90^\circ - \angle BHD = 90^\circ - \angle AHE = \angle DAC$. By AAA similarity we have $\triangle DBH \sim \triangle DAC$. By comparing corresponding sides:

$$\frac {BD}{DH} = \frac {AD}{DC}\implies \frac 3{DH} = \frac 42$$

so it is immediate that $DH = 1.5$.

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Method 3: The most general solution: use Menelaus' Theorem.

We have shown above that $\triangle ABE \cong \triangle CBE$, and thus $AE = EC$.

Applying Menelaus' Theorem on $\triangle ADC$ and the transversal $BHE$, we have

$$\frac {AE}{EC} \times \frac {CB}{BD} \times \frac{DH}{HA} = 1$$

This gives

$$\frac {DH}{HA} = \frac{BD}{CB} \times \frac {EC}{AE} = \frac3{3+2}\times \frac11 = \frac35$$

which is the ratio you seek.