Given: △KLM, KS - angle bisector Prove: KM>SM

Mathematics

1 answer:

Alex787 [66]4 years ago

3 0

Answer:

See below

Step-by-step explanation:

∠KSM = ∠SKL + ∠KLS . . . . . external angle is the sum of opposite internal angles

In order for SM > KM (in contradiction to the hypothesis), we must have

∠MKS > ∠KSM . . . . . . . . . . . angle measures are in the same order as opposite side measures

However, ∠MKS ≅ ∠SKL (given), so SM > KM requires ...

... ∠SKL > ∠SKL + ∠KLS

For any triangle with non-zero area, ∠KLS > 0, so the above statement will be false (a contradiction)

Negating the hypothesis results in a contradiction, so we conclude the hypothesis is true: KM > SM.

You might be interested in

What is the last common denominator of 9/10 and 3/5

Sindrei [870]

5,10,15,20,25,30,35,40

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to (2^5)^-2

irinina [24]

1/1024 is the correct answer. First you had to apply exponent rule, and it gave us, $\frac{1}{(2^5)^2}$ , or $2^5^*^2$ . Then you can also refine, and it gave us, $2^10$ , or $\frac{1}{2^10}$ . And it gave us the answer is $2^1^0=2*2*2*2*2*2*2*2*2*2=1024$ , or 1/1024 is the correct answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie