Similar triangles and altitudes

Question

$\triangle ABC$ is an isosceles $(AC=BC)$. $CM$ and $AA_1$ are altitudes and $CM\cap AA_1=H$. If $CH=12$ and $HM=4$, I should find $AB$.

$CM$ is height in $\triangle ABC$, which is isosceles. Therefore, $M$ is the midpoint of $AB$. Now I am trying to find a relationship between $AM, CH$ and $HM$ but I don't see it. Can you give me a hint? Can we solve the problem using not the particular lengths of $CH$ and $HM$ but $CH=a$ and $HM=b$?

Answer 1

since CH/CM=2/3 so AB=AC and you can use Pythagoras to find AB

Answer 2

For the general case it is better to set$\measuredangle CAB = \measuredangle ABC = \alpha$.

Then, you easily get $$\overline{CH} =\overline{AC} \cdot \sin \alpha$$ $$\overline{AH} = \overline{AC}\cdot \cos\alpha$$ and $$\overline{MH} = \overline{AC}\cdot \frac{\cos^2\alpha}{\sin \alpha}.$$

So the relationship between the angle $\alpha$ and the ratio $r = \frac{\overline{MH}}{\overline{CH}}$ is $$r = \cot^2 \alpha,$$ i.e. (EDIT 1) $$\boxed{\overline{AB} = 2\sqrt r\cdot \overline{CH}}.$$

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EDIT 2

From the above discussion, in your case you have $r=\frac14$ and thus $$AB \cong CH.$$

If you prefer a proof via Euclidean geometry , you might want to inscribe $\triangle ABC$ in the rectangle $ABPQ$, as in the Figure below, where the altitude $AA'$ as been produced to $A''= AA' \cap BP$. Let again $r = \frac{\overline{MH}}{\overline{CH}}$.

1. Intercept Theorem gives you $$\overline{A''B} = 2r \cdot \overline{CH}.$$
2. Similarity $\triangle A'BA''\sim \triangle MA'C$ yields $$\overline{A'A''} = \frac{2r}{1+r}\cdot\overline{MA''},$$ and, since $AM\cong MA''$, $$\overline{A'A''} = \frac{r}{1+r}\cdot\overline{AA''}.$$
3. Euclid on $\triangle ABA''$ gives $$\overline{AA''} =\sqrt {\frac{1+r}{r}}\cdot \overline{A''B}.$$
4. Point 1. and 3., with Pythagorean Theorem leads to the desired result $$\boxed{\overline{AB}=2\sqrt r\cdot\overline{CH}}.$$

Answer 3

Comment: As can be seen in drawing the measures of the base and altitude are equal, therefore AB=CM=16.

You must prove that Altitude is equal to the base.