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

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.

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$