Skip to main content

Visit this link for 1 : 1 LIVE Classes.

For a triangle, show that the point where altitude cuts circumcircle is the image of the orthocentre about the side.


Let ABC be triangle having H as orthocentre. The proof is given below for reflection about side BC. Similar approach can be used for reflections about other sides. 

To prove that HL = PL

Join BP. The idea is to prove that the triangles BLH and BLP are congruent.

Both triangles are right triangles having same base BL.

Arc PC subtends equal angles PBC and PAC.

$\angle PBC = \angle PAC$

From $\Delta ACL$, $\angle LAC = 90^\circ - C$

So, $\angle PBC = \angle PAC =\angle LAC = 90^\circ - C$

From, $\Delta BMC$, $\angle MBC = 90^\circ - C$

So, $\angle PBC = 90^\circ - C = \angle MBC$

From ASA congruence, triangles BLH and BLP are congruent.

From CPCT, HL = PL.

Popular posts from this blog

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
Read more