Triangles ΔABC ≅ ΔBAD so that C and D lie in the opposite semi-planes of segment AB. Prove that segment CD bisects segment AD.
1 answer:
Answer:
See explanation
Step-by-step explanation:
Triangles ΔABC and ΔBAD are congruent. So,
- AB ≅ BA;
- AC ≅ BD;
- BC ≅ AD;
- ∠ABC ≅ ∠BAD;
- ∠BCA ≅ ∠ADB;
- ∠CAB ≅ ∠DBA.
Consider triangles AEC and BED. In these triangles,
- AC ≅ BD;
- ∠EAC ≅ ∠EBD (because ∠CBA ≅ ∠BAD);
- ∠AEC ≅ ∠BED (as vertical angles).
So, ΔAEC ≅ ΔBED. Thus,
AE ≅ EB.
This means that segment CD bisects segment AD.
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Answer:
Value of h greater than 5.4 will make inequality false.
Step-by-step explanation:
This question is incomplete; here is the complete question.
The Jones family has saved a maximum of $750 for their family vacation to the beach. While planning the trip, they determine that the hotel will average $125 a night and tickets for scuba diving are $75. The inequality can be used to determine the number of nights the Jones family could spend at the hotel 750 ≥ 75 + 125h. What value of h does NOT make the inequality true?
From the given question,
Per night expense (expected) = $125
If Jones family stays in the hotel for 'h' days then total expenditure on stay = $125h
Charges for scuba diving = $75
Total charges for stay and scuba diving = $(125h + 75)
Since Jones family has saved $750 for the vacation trip so inequality representing the expenses will be,
125h + 75 ≤ 750
125h ≤ 675
h ≤
h ≤ 5.4
That means number of days for stay should be less than equal to 5.4
and any value of h greater than 5.4 will make the inequality false.
Step-by-step explanation:
Hey, there!!
Given that,
{ we can write (a^2-4) as (a^2 - 2^2) also as (x^2- 9) can be written as (x^2 - 3^2)}.
We have a^2-b^2= (a+b) (a-b), so keep same formula on it.
(x+3) in numerator and denominator gets cancelled,
Therefore, (x-3) is the final value.
<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>