We can see two similar triangles ::
ABC and BFA
Calculate and explain, what is AB/AC?
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khanujarashmit:
Qba: plz specify u want value of AB/AC in terms of triangle BFA ? Or on general u want value of that ratio
Yup, it was a value of that ratio. There was a small mistake though, as KvnMurty noticed, there are more challenging ratios to calculate and I wanted to ask for one of them.
Unfortunately, it was too late to change. AB/AC will always be 1, as was mentioned before :(
Answers
Answered by
22
Given, AB = AC , AD = DB, AE = EC.
So ΔABC is an Isosceles triangle and D and E are midpoints of sides.
Obviously AB / AC = 1 as given. More challenging is the ratio of FD/AB and FE/BC.
Given that ΔABC and ΔABF are similar. On top of that, AB is common between the two. Hence, they are congruent triangles. Their area is same.
As D is the midpoint of AB, FD is the median from F in ΔABF.
Hence, FA = BC, FB = AC = AB.
Using the formula for a median in a triangle (ABF)
4 FD² = 2 FB² + 2 FA² - AB²
= 2 AB² + 2 BC² - AB²
= AB² + 2 BC²
FD = 1/2 * √[AB²+2BC²]
Since DE is a line joining the midpoints of sides, DE = 1/2 BC.
Hence, FE = FD - DE = 1/2 [ √(AB²+2BC²) - BC ]
So ΔABC is an Isosceles triangle and D and E are midpoints of sides.
Obviously AB / AC = 1 as given. More challenging is the ratio of FD/AB and FE/BC.
Given that ΔABC and ΔABF are similar. On top of that, AB is common between the two. Hence, they are congruent triangles. Their area is same.
As D is the midpoint of AB, FD is the median from F in ΔABF.
Hence, FA = BC, FB = AC = AB.
Using the formula for a median in a triangle (ABF)
4 FD² = 2 FB² + 2 FA² - AB²
= 2 AB² + 2 BC² - AB²
= AB² + 2 BC²
FD = 1/2 * √[AB²+2BC²]
Since DE is a line joining the midpoints of sides, DE = 1/2 BC.
Hence, FE = FD - DE = 1/2 [ √(AB²+2BC²) - BC ]
Sir was on half way to solve...but stopped when got ratio 1
Answered by
7
HEYA ❕ HERE'S YOUR ANSWER
Given, AB = AC , AD = DB, AE = EC.
So ΔABC is an Isosceles triangle and D and E are midpoints of sides.
Given that ΔABC and ΔABF are similar. On top of that, AB is common between the two. Hence, they are congruent triangles. Their area is same.
As D is the midpoint of AB, FD is the median from F in ΔABF.
Hence, FA = BC, FB = AC = AB.
Using the formula (ABF)
4 FD² = 2 FB² + 2 FA² - AB²
= 2 AB² + 2 BC² - AB²
= AB² + 2 BC²
FD = 1/2 * √[AB²+2BC²]
Since DE is a line joining the midpoints of sides, DE = 1/2 BC.
Hence, FE = FD - DE = 1/2 [ √(AB²+2BC²) - BC ]
HOPE IT HELPS ✌️
Given, AB = AC , AD = DB, AE = EC.
So ΔABC is an Isosceles triangle and D and E are midpoints of sides.
Given that ΔABC and ΔABF are similar. On top of that, AB is common between the two. Hence, they are congruent triangles. Their area is same.
As D is the midpoint of AB, FD is the median from F in ΔABF.
Hence, FA = BC, FB = AC = AB.
Using the formula (ABF)
4 FD² = 2 FB² + 2 FA² - AB²
= 2 AB² + 2 BC² - AB²
= AB² + 2 BC²
FD = 1/2 * √[AB²+2BC²]
Since DE is a line joining the midpoints of sides, DE = 1/2 BC.
Hence, FE = FD - DE = 1/2 [ √(AB²+2BC²) - BC ]
HOPE IT HELPS ✌️
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