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Question Number 103672 by M±th+et+s last updated on 16/Jul/20

Commented by M±th+et+s last updated on 16/Jul/20

$i f$ $A D + A C = B C$ $a n d$ $A B + B D = A C$ $f i n d ∠ A B C$
Commented by som(math1967) last updated on 16/Jul/20

$If ABC is triangle then how$ $AB + BC = AC ?$
Commented by M±th+et+s last updated on 16/Jul/20

$s o r r y s i r i t s t y p o$ $t h a n k y o u f o r y o u r c o m m e n t$
	Answered by Worm_Tail last updated on 16/Jul/20

	${A C}^{2} = {B C}^{2} + {B A}^{2} - 2 (B C) (B A) c o s (A \hat{B C}) c o s i n e r u l e$ ${(A B + B C)}^{2} = {B C}^{2} + {B A}^{2} - 2 (B C) (B A) c o s (A \hat{B C}) g i v e n$ ${A B}^{2} + 2 (A B) (B C) + {B C}^{2} = {B C}^{2} + {B A}^{2} - 2 (B C) (B A) c o s (A \hat{B C})$ $2 (A B) (B C) = - 2 (B C) (B A) c o s (A \hat{B C})$ $- 1 = c o s (A \hat{B C}) \Rightarrow A \hat{B C} = 180 °$
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