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3 years ago

Given: AC bisects angle BCD , Angle ABC is congruent to angle ADC prove AB is congruent to AD. Help immediately

Mathematics

2 answers:

weeeeeb [17]3 years ago

7 0

AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)

thus in ΔABC and ΔADC, ∠ABC=∠ADC (given),

∠BAC=∠CAD [from (1)],

AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC

Hence, by AAS axiom, ΔABC ≅ ΔADC,

Therefore, BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD), since (1)

Hence, BC=DC proved.

olga55 [171]3 years ago

7 0

Answer:

Solution given;

<BCA=<ACD

<ABC=<ADC

now

In traingle ABC & ∆ ACD

AC=AC[common base]

<BCA=<ACD[bisector bisect it]

<ABC=<ADC[Given]

∆ABC is congruent to ∆ ACD by S.A.A. axiom

AB is congruent to AD.

[ corresponding sides of a congruent triangle are equal]

Hence proved.

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Please don't ask 2 questions in 1 question

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3 years ago

A store sells both cold and hot beverages. Cold beverages, c, cost $1.50, while hot beverages, h, cost $2.00. On Saturday, drink

ra1l [238]

Answer:

The number of hot beverages sold were 45 and the number of cold beverages sold were 180.

Step-by-step explanation:

Let the number of cold beverages sold be $c$

And the number of hot beverages sold be $h$

According to the question:

c=4 $\times$ h...

$$$ 1.5 $\times$ c + 2 $\times$ h = Sale of any day

Part1:

For Saturday.

The sale is of $$$ 360

then

$$$ 1.5 $\times$ c + $$$ 2 $\times$ h = $$$ 360

Part 2:

Following the method of substitution.

And plugging the values of c=4 $\times$ h...in equation where the sales of Saturday is given.

1.5 $\times$ c + 2 $\times$ h =360

1.5 $\times$ 4 $\times$ h + 2 $\times$ h =360

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Dividing both sides with 8.

h= $\frac{360}{8}$

h=45

Inserting h=45 in c=4 $\times$ h...

we have c=4 $\times$ 45 = 180

So the number of hot beverages sold were 45 and the number of cold beverages sold were 180.

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