ABC-is-a-triangle-such-that-AB-AC-D-is-a-point-on-side-AC-such-that-BC-2-AC-CD-Prove-that-BD-BC-

Question Number 214859 by MATHEMATICSAM last updated on 21/Dec/24

ABC is a triangle such that AB = AC .

D is a point on side AC such

that {BC}^{2} = AC . CD . Prove that

BD = BC .

Answered by A5T last updated on 21/Dec/24

Commented by A5T last updated on 21/Dec/24

{B C}^{2} = C A \times C D \Rightarrow C D = \frac{{B C}^{2}}{C A}

∠ B C D = θ \Rightarrow {A B}^{2} = {B C}^{2} + {A C}^{2} - 2 B C \times A C c o s θ

\Rightarrow c o s θ = \frac{B C}{2 A C}

{B D}^{2} = {B C}^{2} + {C D}^{2} - 2 B C \times C D c o s θ

\Rightarrow {B D}^{2} = {B C}^{2} + {C D}^{2} - C D \times \frac{{B C}^{2}}{C A}

\Rightarrow {B D}^{2} = {B C}^{2} + {C D}^{2} - {C D}^{2} = {B C}^{2}

\Rightarrow B D = B C

Commented by MATHEMATICSAM last updated on 22/Dec/24

Can it be proved by pythagoras

identity without cosine law or something ?

Commented by A5T last updated on 22/Dec/24

I {d o n}^{'} t r e a l l y k n o w a b o u t t h a t, b u t c o s i n e r u l e

c o u l d b e s e e n a s a g e n e r a l i z e d f o r m o f

{P y t h a g o r a s}^{'} t h e o r e m (a s p e c i a l c a s e w h e n θ = 90 °)

Answered by mr W last updated on 22/Dec/24

Commented by mr W last updated on 22/Dec/24

{B C}^{2} = A C \times C D

\Rightarrow \frac{B C}{A C} = \frac{C D}{B C}

\Rightarrow Δ A B C \sim Δ B D C

\Rightarrow ∠ B D C = ∠ A B C = ∠ B C D

\Rightarrow B D = B C

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