Whther median to hypotenuse in Right-angled ∆ is circumradius of ∆?

Question

I have read following points in a quant book but am not able to understand them. Can anyone help me solving my belowmentioned queries:
1. Median to hypotenuse in Right-angled ∆ is circumradius of ∆. How??
2. Area of ∆ = rXs = abc/4R where s = semi perimeter, R = circumradius & r = inradius...... Can anyone provide me proof of the same.
3. Whether in an isosceles triangle inradius = 1/3 of height??
4. In chapter of trignometry it is stated that Area of ∆ = 1/2 X Product of 2 sides X sine of included angle... As per me it is applicable only in case of Right angle in a Right Angled ∆ & to no other angle.

You all are heartly requested to help me in resolving my queries.
Rgds
Parag

Sorry I didn't get you as per Sine rule a/sin A=b/sin B = c/sin C but how you have taken it equivalent to 2R.
Further for question no. 3 you have written Saludos... which I was unable to comprehend

Thanks Rick for your answer no. 1 but could you please help me in solving question 4 if:
AB = 3, AC =5 & Angle ABC is 90 & Angle BAC is 30
Now as per the formula area = 1/2 X 3 X 5 X sin (30) = 15/4
though if we calculate BC via pythagorus which comes on to 4 & calculate area now it comes on to = 1/2 X 3 X 4 X sin (90) = 6

Please help me in correcting my mistake.

Answer 1

1) T. seno: a/sinA = b/sinB = h/sin90º = 2R
--> h= 2·R --> R = h/2 in any hypotenuse in Right-angled ∆

2) Area = r · s because Area = a·r/2 + b·r/2 + c·r/2

and Area = a·b·sinC/2 = a·b· (c/2R) /2 = a·b·c/(4R)

3) A=(2a+b)/2 · r = a^2 · sinC /2 = a · [a·sinC/2] /2 =
a·h /2 --> (2a+b)·r=ah --> (2a+b)·h/3 =ah -->
(2a+b)h=3ah --> a=b --> a=b=c= Equ.

4) Applicable in any angles.

Saludos.

Answer 2

1. Whenever you circumscribe a circle around a right triangle, the hypotenuse always forms a diameter of the circle. This explains why half the hypotenuse is a radius.

2. This sounds like Hero's formula. There are probably many proofs on the Web.

3. Don't know.

4. To see why this is so:
* Draw arbitrary non-right triangle, oriented so that one side is horizontal and at the bottom. This is the "base".
* Extend a vertical line segment downward, from the upper vertex to the line defined by the base. (if the vertical segment does not intersect the triangle's base, extend the base line so that it touches the vertical segment.
* You will now see that you have two right triangles. You can find the area of your original triangle by (depending on the case) either adding or subtracting the areas of the two right triangles. You will find, with algebra, that the resulting area equals (1/2 X Product of 2 sides X sine of included angle).