A line segment BK is an angle bisector of ΔABC. A line KM intersects side BC such, that BM = MK. Prove: KM ∥ AB.
2 answers:
Answer:
∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and lies between AB and KM and BK is the transversal line)
m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)
Step-by-step explanation:
Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MK
TO prove: KM ║AB
Now, As given in figure 1,
In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector)
Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)
Now ∵ ∠MBK = ∠BKM
and ∠ABK = ∠KBM
∴ ∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line)
Hence proved.
You might be interested in
In statistics, mean is the average in a set of data collected. To calculate the mean, you have to add up all the numbers in the table and divide it by the number of data that exists. For example, if the data set is 2 4 1 5 2 8, the mean will be:
mean = = 3.6
So, for this data set, the mean is 3.6.
The median is also a average number, however, the median is the middle value in the list. For example, with the data set above, the median is 3, because, as the set has an even number of data, the middle term would be the mean of the two middle number, in other words:
median = = 3
The standard deviation (σ) is the spread of data distribution, which means it's how "far" a number is from the mean of the set. The formula is
σ = √∑ (x - μ)²/N, where ∑ is the total sum; μ is the mean; and N is the number of data points in the sample. So, to calculate the standard deviation, using the example:
1) Calculate the mean of the set: μ=3.6
2) Find the difference between each data point and the mean and then, the square of each one:
(2-3.6)² = 2.56;
(4-3.6)² = 0.16
(1-3.6)² = 6.76
(5-3.6)² = 1.96
(2-3.6)² = 2.56
(8-3.6)² = 19.36
3) Add up all the squares: ∑= 33.36
4) Divide by the number of data points: =5.56
5) Take the square root: σ = 2.36
For the set of numbers above (2,4,1,5,2,8), the standard deviation is
σ = 2.36.