Question 2:
$\displaystyle \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}\Ri ghtarrow 2(x+y+z)=180^{\circ}\Rightarrow x+y+z=90^{\circ}$
$\displaystyle \widehat{ADB}=180^{\circ}-2z=180^{\circ}-2(90^{\circ}-(x+y))=$
$\displaystyle =180^{\circ}-180^{\circ}+2(x+y)=2\widehat{ACB}$
Question 4:
$\displaystyle \widehat{FBG}$ is exterior to triangle $\displaystyle BDG$. So $\displaystyle \widehat{FBG}>\widehat{BDG}$ (1).
$\displaystyle \widehat{BDG}$ is exterior to triangle $\displaystyle ADE$. So $\displaystyle \widehat{BDG}>\widehat{AED}$ (2).
From (1) and (2) via tranzitivity, yields $\displaystyle \widehat{FBC}>\widehat{AED}$
Hello,
to question 2:
I've modified your drawing (see attachment):
The angle in red is an exterior angle of the isosceles triangle ACD;
the angle in blue is an exterior angle of the isosceles triangle ABD.
The angle at the centre is 2x + 2y = 2(x+y) that means it is twice as large as the angle at C.
Hello, Raiden_11!
Basically the same as red_dog . . .
In $\displaystyle \Delta ABC\!:\;2x + 2y + 2z \;=\;180^o\quad\Rightarrow\quad x + y + z \;=\;90^o$
. . $\displaystyle \angle BCA \;=\;x + y \;=\;90^o - z$
In $\displaystyle \Delta ADB\!:\;\angle ADB + 2z \;=\;180^o$
Therefore: .$\displaystyle \angle ADB \;=\;180 - 2z \;=\;2(90^o-z) \;=\;2\cdot\angle BCA$
2. Observe the relationship between the elements of Diagonal 1 and Diagonal 2 in Pascal’s Triangle, shown on the next page. The sum of the first two elements in Diagonal 1 is equal to the second element in Diagonal 2. Similarly, the sum of the first 5 elements of Diagonal 1 is equal to the 5th element of Diagonal 2. Suppose we wish to use Pascal’s Triangle to find the sum of the first n natural numbers. Use combinatorial notation to express the sum of the n elements of Diagonal 1 in terms of the corresponding value in Diagonal 2.
3. Using the method of mathematical induction, prove that the following statements are false.
2.
$\displaystyle C_1^1+C_2^1=C_3^2$
$\displaystyle C_1^1+C_2^1+C_3^1+C_4^1+C_5^1=C_6^2$
$\displaystyle C_1^1+C_2^1+\ldots C_n^1=C_{n+1}^2$
3.
For the first identity:
For $\displaystyle n=1\Rightarrow 2=2$
For $\displaystyle n=2\Rightarrow 6=6$
For $\displaystyle n=3\Rightarrow 12=14$, false. So the identity is not true for all $\displaystyle n\in\mathbf{N^*}$.
For the second identity:
For $\displaystyle n=1\Rightarrow 2=4$, false. So the identity is not true for all $\displaystyle n\in\mathbf{N^*}$.