 Hello and welcome to another session on gems of geometry as We have discussed many important concepts. So furthering that pursuit. We are going to discuss another one now So this particular theorem says that if circles are constructed on two CVN says diameters Their radical axis passes through the ortho center edge of the triangle now the prerequisite for this session would be the concept of CVN, you know that a CVN is nothing but a line joining any of the vertex to any of the points on the other side of The vertex in a triangle. That's the CVN So in the adjoining figure you can say BE, AD and CF are CVNs It's another fact that they are a special type of CVN which is called an altitude because they are perpendicular to the sides here Then there's another concept of radical axis. What is radical axis before? Getting into radical axis. We had discussed something called power of a point with respect to a circle So if you remember Or if you don't really if you haven't been through those sessions, I would urge you can go back to those sessions where we have discussed what does a power of a point with respect to a circle mean But just to give you a quick recap the power of a circle, let's say a point P Inside or outside the circle is nothing but a d square minus r square Where d is nothing but a distance between distance between point P and center Okay, and r is nothing but the radius of the circle radius of the same circle So this is called power of a point with respect to circle and we also saw That the power of a point with respect to circle is nothing but for example, you know Let me it's it's better to take an example and illustrate. So let's say we have a point P here Let's say this is P. So power of a point P here in this case Would be nothing but let's say if I join the Let's say a chord passing through P Okay, so we had discussed in the previous session. We are not going to discuss the proof once again I would request if you don't know this concept You can have a view of the previous sessions where we have discussed the power of a point now power of a point P Let's say this is a and this is let's say a dash. So power for point is also defined as P is equal to Power sorry power power of the point is equal to a P into P a dash Okay, so let's say P is lying on a card then A and a dash are the points on the circumference or a a dash is the card Then power is also defined power of which point P is defined as a P into P a dash This was the information now in this case if you see H happens to be the ortho center here Ortho center is nothing but point of concurrence of all the three altitudes of a triangle Ortho center H is the ortho center. So if H is the ortho center in H happens to be let's say If I want to find out the power of point H with respect to a given circle then H must lie on one of the chords and then This kind of relationship will give you the power of H. Okay, so we'll park this discussion for some for some time Let's now understand You know another another important fact or another important property of The ortho center now if you see H is the ortho center and now they are talking about a sevian. Let me just remove all these points So they are talking about the sevian. Let's draw a sevian guys. So if I draw from a let's say a Let's say a P is the sevian. Okay. AP is the sevian now if AP is the sevian and if you draw a circle Using AP as the diameter then clearly point D will lie on that circle, isn't it? It will be something like that. So roughly drawing like that, right? So why would D lie on the circle because AP is the diameter and if you see ADP So let me write that angle ADP is 90 degrees isn't it so if ADP is 90 degrees and AP is the diameter then Clearly a D must lie on the semi circle. So right why because in a semi circle the angle subtended by a diameter is 90 degrees so hence we can claim that D lies D lies on the circle on the circle with AP as diameter Isn't it? We can say that now Can we now if you see now H happens to be H is now inside the circle H is a point H is a point inside Inside or even if it is outside doesn't matter. So H is a point inside outside. Let's say outside right Of a circle Of a circle with dia with Diameter how much AP Meaning thereby I can find out the power of point H. So power of point H power of Power of H will be how much as per the previous discussion So ad happens to be the chord. So hence I can say power of H is ah Power of H is nothing but ah into h D Isn't it? Yep ah into hd similarly and so hence what is it? So ap is a cvn and using ap as a cvn as a diameter If you draw a circle then the power of point H with respect to that circle will be ah into hd So now come let us consider another cvn from let's say point b So if you draw another cvn, let's say this is another cvn and let's call it q bq bq the cvn Now if you see bq is a cvn then clearly e lies on that circle e lies on that circle which Right where bq is a diameter. Why again the same logic e happens to be lying on a circle Whose diameter is bq correct? Why because the angle subtended by a diameter on the semi circle or the circle is 90 degree And b e happens to be 90 degree here, right? So hence clearly e lies on that circle So we can find out the power of point h with respect to That circle this new circle also right whose diameter is bq correct? So in that case what will be the power of h in that case power of h power of h With respect to let's say i'm naming this as bq bq is the diameter. So in this case it was ah ap Right so ap was the diameter here power of h with respect to bq will be clearly equal to E h or b h into h e, isn't it? dh into he no problems, right? So this is what we can clearly Find out. Yeah, so power of H with respect to ap is this power of h with respect to bq is this now they're saying that There are two circles one with You know Diameter ap and another circle with diameter bq Now what is meant by radical axis? So radical axis is nothing but if you remember in the previous session you can check a radical axis is radical axis Is nothing but locus of locus of all points whose power with respect to the given circles is always equal or are always Equal that means if I find out the power of a point with respect to circle one And find out the power of the circle with respect to circle two then both of them will be equal Now the theorem claims that if you draw two such circles with two civilians as diameter then the Radical axis will pass through h that is h lies on the radical axis of the two Circles right that means the power of h with respect to the first circle as well as power of h with respect to the second circle Must be equal Which means This must be equal to this which which means if we somehow prove that a h into hd Is equal to b h into he Then our job is done If somehow we prove that this is true then mean the power of Power of h with respect to both the circles are equal that means Whatever is the radical axis of the two given circles h would be one point of them on them, isn't it? One point on that radical axis so the point of the entire problem is reduced to proving this so let's say if you can prove Ah into hd is equal to bh into he or not once again just to reiterate and re-emphasize once more So power of h with respect to the first circle is this a h into hd Power of h with respect to the second circle is bh into he Now the theorem claims that h lies on the radical axis of both the circles And what is the radical axis of the two circles nothing but a locus of all the points Uh such that the power of the point with respect to both the circles are equal Now if h happens to be on the radical axis that means the power of h with respect to the first circle Must be equal to power of h with respect to the second circle That means we have to prove that a h into hd is equal to bh into he that is this is what we have to achieve Let's say we let's find out if we can really achieve that Okay So how to achieve that so it will be very simple to see again. Let me do it like that. Yeah, so now consider Uh Angle b da Is equal to angle b e a b e a b da is equal to angle b e a and both of them are equal to 90 degrees. That means what can we say if We draw If we draw a circle Circle with a b as the diameter a b as the Diameter diameter then then point d and e would would lie on the Circumference is it D and e will lie on the circumference Now in a given circle, we know that let's say there is a two there's a circle. Okay Let's say there is a circle There's a circle If you draw two cards here We had you know again proven this in the previous sessions. Let's say the two cards are there a b and c d Okay, then we had proved that a let's say this is p So ap into pb is equal to cp into pd Right by similar Triangles theorem we can Prove that and we have done that you can check it check it out in the previous session Okay, now if that is so then if you consider a circle passing through Or a with a b as diameter and passing through d and e then clearly h Is the point of intersection of two cards which two cards? So two cards are ad and Ae are the two cards Are two cards? Correct two cards intersecting At edge Right therefore what do we Infer so that means bh into he Will be equal to ah into hd By this theorem the proof of which we can check out in the previous sessions So hence if you see this is actually true guys, so if you see this is what we needed to prove, isn't it? So bh into he is equal to ah into hd. This is what we needed to prove and it actually true It's actually true. Therefore we can conclude conclusion. So what is the conclusion? Conclusion is clearly edge lies or power of we can say power of power of edge with respect to with respect to both circles with respect to circles with CVNs ap and b q as diameter as diameters are equal Power of edge with respect to circles with CVNs ap and b q as diameters are equal. That means this implies edge lies on the radical axis of both the circles Right. This is what we needed to prove and hence proved I hope you understood the theorem interesting theorem talks about the radical axis passing through the orthocenter of um radical axis of which two circles circles made On any two CVNs as diameter will pass through the orthocenter. Okay