ABC is a triangle in which the bisector of angle at B meet the side AC at D, and the bisector of the angle BDC is parallel to the side AB. Prove that the △ABC is issoceles triangle.


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Question Number 169145 by MathsFan last updated on 24/Apr/22

$A B C is a triangle in which the bisector$ $of angle at B meet the side A C at D,$ $and the bisector of the angle B D C is$ $parallel to the side A B . Prove that$ $the △ A B C is issoceles triangle .$
	Answered by som(math1967) last updated on 25/Apr/22

	$l e t b i s e c t o r o f ∠ B D C m e e t t h e$ $s i d e B C a t E$ $∠ A B D = ∠ D B E = α [B D b i s e c t o r]$ $∠ A B D = ∠ B D E [a l t e r n a t e]$ $∠ B D E = ∠ E D C = α [D E b i s e c t o r]$ $A B ∥ D E ∴ ∠ E D C = ∠ B A C = α$ $A C = b, A B = c, B C = a$ $∴ \frac{a}{s i n α} = \frac{b}{s i n 2 α}$ $\Rightarrow \frac{a}{s i n α} = \frac{b}{2 s i n α c o s α}$ $\Rightarrow c o s α = \frac{b}{2 a}$ $\Rightarrow \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{b}{2 a}$ $\Rightarrow b^{2} a + c^{2} a - a^{3} - b^{2} c = 0$ $\Rightarrow b^{2} (a - c) - a (a^{2} - c^{2}) = 0$ $\Rightarrow (a - c) (b^{2} - a c - a^{2}) = 0$ $a - c = 0 \Rightarrow a = c$ $t r i a n g l e i s i s o s c a l e$
	Commented by som(math1967) last updated on 25/Apr/22

	Commented by som(math1967) last updated on 25/Apr/22

	$a n y o t h e r m e t h o d w i t h o u t$ $t r i g o n o m e t r y ?$
	Commented by MathsFan last updated on 25/Apr/22

	$t h a n k y o u s i r$
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