Question

traingle ABC, if 2 C > 2 B, then,
BC> AC
AB> AC
(c) AB<AC
(d) BC<AC​

Answer 1

Answer:

Step-by-step explanation:

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]

hope it helps

:)