Heya! ツ ☛ : In the given figure , O is the centre of the circle. Two equal chords AB and CD intersect each other at E. Prove that :
i. AE = CE
ii. BE = DE

*Random/irrelevant answers will not be tolerated !

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:

this is your answer. look it once.

2nd one in the first row is correct

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,