Answer:
A
Step-by-step explanation:
Answer:
See Below.
Step-by-step explanation:
In the given figure, O is the center of the circle. Two equal chords AB and CD intersect each other at E.
We want to prove that I) AE = CE and II) BE = DE
First, we will construct two triangles by constructing segments AD and CB. This is shown in Figure 1.
Recall that congruent chords have congruent arcs. Since chords AB ≅ CD, their respective arcs are also congruent:

Arc AB is the sum of Arcs AD and DB:

Likewise, Arc CD is the sum of Arcs CB and DB. So:

Since Arc AB ≅ Arc CD:

Solve:

The converse tells us that congruent arcs have congruent chords. Thus:

Note that both ∠ADC and ∠CBA intercept the same arc Arc AC. Therefore:

Additionally:

Since they are vertical angles.
Thus:

By AAS.
Then by CPCTC:

Let 'x' be the distance from THE far bank where 700 is the distance to the NEAR bank
boat one has travelled 700 (rate = 700/unit time) boat two has travelled x rate = x / unit time
boat one then travels x + 400 more and boat two travels 700 + (700+x -400) more when they meet
The time is the same rate x time = distance distance/rate = time equate the distances divided by the respective rates
(700 + x + 400)/700 = ( x + 700 + (700+x-400) )/x
1100x + x^2 = 1400x + 700000
x^2-300x -700000 = 0 quadratic formula yields x = 1000
One boat travels 700 the other 1000 whe they first meet.....width of river = 700+ 1000 = 1700 m
Since these are supplementary angles (two angles which sum to 180°) we can say:
4x+x=180 combine like terms on left side
5x=180 divide both sides by 5
x=36°