Answer:
There's no pic, mate
Step-by-step explanation:
Answer:
see below
Step-by-step explanation:
<h3>Proposition:</h3>
Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE
<h3>Proof:</h3>
1)The lines AD and BC are parallel and AC their transversal therefore,
![\displaystyle \angle DAC = \angle ACB \\ \ \qquad [\text{ alternate angles theorem}]](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cangle%20DAC%20%3D%20%20%5Cangle%20ACB%20%5C%5C%20%20%5C%20%5Cqquad%20%5B%5Ctext%7B%20alternate%20angles%20theorem%7D%5D)
2)The lines AB and DC are parallel and BD their transversal therefore,
![\displaystyle \angle BD C= \angle ABD \\ \ \qquad [\text{ alternate angles theorem}]](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cangle%20BD%20C%3D%20%20%5Cangle%20ABD%20%5C%5C%20%20%5C%20%5Cqquad%20%5B%5Ctext%7B%20alternate%20angles%20theorem%7D%5D)
3)now in triangle ∆AEB and ∆CED
therefore,

hence,
Proven
Answer:
a = 82
b = 130
c = 118
Step-by-step explanation:
A quadrilateral is inscribed in a circle. So, it is a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary.
Therefore,
a° + 98° = 180°
a° = 180° - 98°
a° = 82°
a = 82
By inscribed angle theorem:
a° = 1/2(b° + 34°)
82° * 2 = b° + 34°
164° - 34° = b°
b° = 130°
b = 130
Again by inscribed angle theorem:
76° = 1/2(c° + 34°)
76° *2 = c° + 34°
152° - 34° = c°
c° = 118°
c = 118
Answer:
4a+1/3a=14
13a/3=14
a=42/13
Step-by-step explanation: