in the adjoining figure, show that EI is the bisector of <KET and <KIT....... get 100 points for free by answering the correct answer !!!
Answers
Step-by-step explanation:
In triangle ABC, ∠B = 90°
seg BD ⊥ Hypotenuse AC
∴ ΔABC ≈ ΔADB
#(13 * 5 = 65)
#Refer the figure for better understanding.
Take the length of AC = 13 c.m and and mark point D at 5 c.m.
∴ AB/AD = AC/AB
∴ AB² = AD × AC
∴ AB² = 13 × 5
∴ AB² = 65
∴ AB = √65
So, Length of the seg AB is √65.
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