A Nice geometry math question| find the area of the triangle.

You're welcome back to my channel. I'm super excited to solve this trigonometry geometry question with you. Please this is not drawn to scale and in this question we are asked to find the value of the area of this triangle. This is a right angle triangle. If from A to this point, let's call this point P, okay, is 3 and from P to C is 4. The question says we should find the area of this triangle. Okay, first of all let's remember the formula for area of a triangle. Now area of triangle ABC should be equal to half base. The base is BC multiplied by the height and the height is AB. So we are going to try as much as possible to find the value of BC and the value of AB. Alright, so first of all I want us to remember this theorem because it's very important to us right now. Now this theorem says that from an extended point the length of two tangents are equal. From an extended point the lengths of two tangents are equal. Now what this means is that if we have a circle, okay, let's call this if this is a circle because it's a freehand sketch. Now if you have a circle and you have two tangents, this is the first tangent, this is the second tangent and they meet at point P. So AP and BP are equal. That is what this particular theorem is talking about. So using this particular theorem let's go back to our diagram. Now you can see that AP and let's call this particular point, point C, point D, sorry we already have C and call this particular point, point E, okay. Now you can see that AD and AP are tangents to this circle and their meeting point is A. So it simply means that AP should be equal to AD, right. So it means that AD is also equal to three. So let's write that. Now also let us consider PC and EC, right. So PC and EC are tangents to this particular circle at points P and E and they all meet at point C. So it simply means that PC should be equal to EC which is equal to four. All right now we can go ahead and look at point B as well. You can see that DB and BE are tangents to this circle at points D and E and they meet at point B. So it simply means that DB is equal to BE, right. So if we call this place Y and this place Y as well because they should have equal length. Now let us say BC minus AB, okay. So BC minus AB is equal to, now BC is Y plus four. So we have Y plus four minus AB is three plus Y or Y plus three. So we have Y plus three, okay. Now I'm trying to adjust this to be comfortable. Okay now we have BC minus AB is equal to Y plus four minus times Y is minus Y minus times plus three is minus three. So we have BC minus AB is equal to Y minus Y plus four minus three. If we collect like terms. So BC minus AB is equal to Y minus Y is zero and four minus three is one. Okay now let's square both sides. Squaring both sides we have BC minus AB all squared is equal to one squared. Now remember that A minus B all squared is equal to A squared plus B squared minus two AB. So it means that BC minus AB squared should be equal to BC squared plus AB squared minus two multiplied by BC multiplied by AB is equal to one squared is one. Now I want us to look at this diagram again. In triangle ABC, remember that triangle ABC is a right angle triangle, okay. Now because it's a right angle triangle it simply means that AB squared plus BC squared is equal to AC squared, okay. So but in triangle ABC using Pythagorean theorem AB squared plus BC squared is equal to AC squared. So we have AB squared plus BC squared is equal to now AC is what three plus four right so is equal to three plus four squared. So we have AB squared plus BC squared is equal to seven squared. So AB squared plus BC squared is equal to 49. Now we go back to this particular equation. So we have BC squared plus AB squared. Remember that that is equal to 49, right. So we are going to replace BC squared