I proved this property with an approach involving vectors. However, there should be a much simpler, elegant geometric proof, probably utilising a bunch of angles. Here is a diagram exemplifying the property:

Orthocentre-Incentre property

And here is my very tedious approach:

Let $ \lambda_1 = \frac{ \overrightarrow{OA} \cdot \overrightarrow{OB}}{\overrightarrow{OA} \cdot \overrightarrow{OA}} = \frac{|\overrightarrow{OB}|}{|\overrightarrow{OA}|} \cos(\theta) $ and $ \lambda_2 = \frac{ \overrightarrow{OA} \cdot \overrightarrow{OB}}{\overrightarrow{OB} \cdot \overrightarrow{OB}} = \frac{|\overrightarrow{OA}|}{|\overrightarrow{OB}|} \cos(\theta) $

By considering the intersection of the lines $BP_1$ and $AP_2$, where

$$ AP_2 := (1-\mu_1)\lambda_1 \overrightarrow{OA} + \mu_1 \overrightarrow{OB} \\ BP_1 := (1-\mu_2)\lambda_2 \overrightarrow{OB} + \mu_2 \overrightarrow{OA} \\ (1-\mu_1)\lambda_1 \overrightarrow{OA} + \mu_1 \overrightarrow{OB} = (1-\mu_2)\lambda_2 \overrightarrow{OB} + \mu_2 \overrightarrow{OA} \\ \implies ((1-\mu_1)\lambda_1 - \mu_2 )\overrightarrow{OA} = ((1-\mu_2)\lambda_2 - \mu_1 )\overrightarrow{OB} \\ \implies (1-\mu_1)\lambda_1 = \mu_2 \quad \text{ and } \quad (1-\mu_2)\lambda_2 = \mu_1 \\ \implies \mu_1 = \frac{\lambda_2 - \lambda_1\lambda_2}{1-\lambda_1\lambda_2} \quad \text{ and } \quad \mu_2 = \frac{\lambda_1 - \lambda_1\lambda_2}{1-\lambda_1\lambda_2} \\ \ \\ \ \\ \implies \overrightarrow{OX} = \frac{1}{1-\lambda_1\lambda_2} \left( (\lambda_1 - \lambda_1\lambda_2)\overrightarrow{OA} + (\lambda_2 -\lambda_1\lambda_2) \overrightarrow{OB} \right) $$

Similarly, by considering the intersection of $AB$ and $OX$,

$$ \overrightarrow{OP_3} = \frac{1}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( (\lambda_1 - \lambda_1\lambda_2)\overrightarrow{OA} + (\lambda_2 -\lambda_1\lambda_2)\overrightarrow{OB} \right) \\ \overrightarrow{P_1P_3} = \frac{1-\lambda_1}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( (\lambda_1 - 2\lambda_1\lambda_2)\overrightarrow{OA} + \lambda_2\overrightarrow{OB} \right) \\ \overrightarrow{P_2P_3} = \frac{1-\lambda_2}{\lambda_1 + \lambda_2 - 2\lambda_1\lambda_2} \left( \lambda_1\overrightarrow{OA} + (\lambda_2 - 2\lambda_1\lambda_2)\overrightarrow{OB} \right) \\ $$

Since we are looking for angles, the magnitudes of the vectors do not matter; we may strip the constants away to simplify our dot product.

$$ \overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_1P_{3\text{ dir}}} = |\overrightarrow{OP_{3\text{ dir}}}| |\overrightarrow{P_1P_{3\text{ dir}}}| \cos(\alpha_1) = \ldots = \sin^2(\theta)\cos^2(\theta) (\overrightarrow{OA} - \overrightarrow{OB})^2 \\ \overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_2P_{3\text{ dir}}} = |\overrightarrow{OP_{3\text{ dir}}}| |\overrightarrow{P_2P_{3\text{ dir}}}| \cos(\alpha_2) = \ldots = \sin^2(\theta)\cos^2(\theta) (\overrightarrow{OA} - \overrightarrow{OB})^2 \\ \implies \frac{\overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_1P_{3\text{ dir}}}}{\overrightarrow{OP_{3\text{ dir}}} \cdot \overrightarrow{P_2P_{3\text{ dir}}}} = \frac{|\overrightarrow{P_1P_{3\text{ dir}}}|\cos(\alpha_1)}{|\overrightarrow{P_2P_{3\text{ dir}}}|\cos(\alpha_2)} = 1 $$

Finally, by expansion, note that $\frac{|\overrightarrow{P_1P_{3\text{ dir}}}|}{|\overrightarrow{P_2P_{3\text{ dir}}}|} = \frac{\left| |\overrightarrow{OB}| \widehat{OA} - 2\overrightarrow{OA}\cos(\theta) + |\overrightarrow{OA}| \widehat{OB}\right|}{ \left| |\overrightarrow{OB}| \widehat{OA} - 2\overrightarrow{OB}\cos(\theta) + |\overrightarrow{OA} \widehat{OB}\right|} $, which can then be geometrically proved to be $\frac{|\overrightarrow{OA}-\overrightarrow{OB}|}{|\overrightarrow{OA}-\overrightarrow{OB}|} = 1$, here is another diagram:

Proving a ratio to be 1

Therefore, $\cos(\alpha_1) = \cos(\alpha_2) \implies \alpha_1=\alpha_2 $.

Now you see why I think there must be a much easier method. How else can we prove this?

  • 1
    $\begingroup$ It's helpful to remember —and easy to prove— that joining the feet of perpendiculars creates a triangle similar to the original. So, $\triangle OAB \sim \triangle OP_2P_1 \sim \triangle P_3AP_1\sim \triangle P_3 B P_2$. Consequently, $\cong \angle AP_3 P_1 \cong \angle O \cong \angle BP_3 P_2$, and we deduce $\alpha_1=\alpha_2$. Etc. $\endgroup$ – Blue Jun 6 '18 at 20:16

$OP_1XP_2$ and $P_1AP_3X$ are cyclic so $$\angle P_2P_1X = \angle P_2OX= \angle P_2AB =\angle P_3P_1X $$ so $P_1B$ is angle bisector for $\angle P_3P_1P_2$...

  • $\begingroup$ $\angle P_2P_1X=\angle P_2OX=90^{\circ}-\angle P_2XO=90^{\circ}-\angle AXP_3=\angle P_3AX=\angle P_3P_1X$. $\endgroup$ – Rosie F Jan 20 '19 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.