solve question 9 plzzzzzzzzzzz
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If AB=EF ,
And ∆FCB and ∆EBC share same base an lie between same parallels.
∆FCB≈∆EBC
By CPCT,
Angle FBC=Angle ECB.
Therefore,
∆ABC is isoceles. By Isoceles triangle property.
And ∆FCB and ∆EBC share same base an lie between same parallels.
∆FCB≈∆EBC
By CPCT,
Angle FBC=Angle ECB.
Therefore,
∆ABC is isoceles. By Isoceles triangle property.
adityasinghyadav77:
its wrong my dear
in maths my friend
in maths there are many ways to solve a question
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GIVEN :-
BE is altitude,
So, Angle AEB = Angle CEB = 90°
CF is altitude,
So, Angle AFC = Angle BFC = 90°
Also, BE = CF
To Prove :-
∆ ABC is Isoceles
PROOF :-
In ∆ BCF and ∆ CBE
Angle BFC = Angle CBE = 90° [ Both angles are 90°]
BC = CB [ Common ]
FC = EB [ Given ]
∆ BCF ≈ ∆ CBE [ RHS Congruency Rule ]
Therefore,
Angle FBC = Angle ECB [ c.p.c.t.]
So, Angle ABC = Angle ACB
AB = AC [ Side Opposite to Equal Angles is Equal ]
So,
∆ ABC IS ISOCELES
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