In AABC", AC² = AB + BC then LB=
Answers
Step-by-step explanation:
In any triangle angle opposite to the longest side is largest.
Angle opposite to AB is ∠C and angle opposite to BC is A
Here AB>AC
⇒∠C>∠A
So option C is correct.
Answer:
In a triangle ABC, angle A = 2 angle B. How would you prove that BC² = AC² + AB × AC?
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I can’t understand why everyone complicates this question. It can be easily solved by similar triangles.
In this png, we have something to make sure.
∠B=∠DAB (Yes, dab)
This also means AD=BD .
This is our basic construction of D, which is going to help us.
∠ADC=∠DAB+∠B=2∠B=∠CAB
∠CAD=∠CAB−∠DAB=∠B
These are based on the fact that ∠A=2×∠B
Actually these conditions suffice. Because I am just proving that △ACD∼△BCA
Similarity makes us realize the following:
ACBC=ADAB
and
ACBC=
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GIVEN: Triangle ABC, <B = x, <A = 2x
TO PROVE THAT: BC² = AC² + AB * AC
CONSTRUCTION: CX perpendicular to AB.
& construct CD = CA ……….. (1)
PROOF: <A = <D = 2x
& in triangle CDB, exterior angle 2x = < B+<C
=> < B = < BCD = x
=> CD = DB ( Sides opposite to equal angles of a triangle) ……….. (2)
Hence, AC = DB …….. (3) By (1) & (2) ………(3)●
Now, by Extension of Pythagoras theorem for acute triangle…
BC² = AC² + AB² - 2AB * AX
=> BC² = AC² + AB ( AB - 2AX)
=> BC² = AC² + AB * ( AB - AD) ( since AX = XD)
=> BC² = AC² + AB * DB
But DB = AC ( by eq (3) )
Hence, BC² = AC² + AB * AC
[ Hence Proved]
Step-by-step explanation:
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