Question

In the given figure AE=AC angle BAC=40 angle ACF=75 and BCF is a line. Prove that BE=CE

Answer 1

Answer:

In ∆ AEC,

AE = AC [given]

∴ ∠AEC = ∠ACE ……. (i) [∵ angles opposite to equal sides are equal]

By Angle sum property,

∠EAC + ∠AEC + ∠ACE = 180°

⇒ 2 * ∠AEC = 180° - 40° = 140° ….. [from (i)]

⇒ ∠AEC = 140° / 2 = 70°

∴ ∠AEC = ∠ACE = 70° ….. (ii)

We are given that BCF is a line, therefore

∠ACF + ∠ACE + ∠ECB = 180°

or,  75° + 70° + ∠ECB = 180° …… [∵ ∠ACF is given 75° and  from (ii) ∠ACE = 70°]

or, ∠ECB = 180° - 145° = 35° ….. (iii)

Now, by using the exterior angle property of a triangle, we have

∠AEC = ∠EBC + ∠ECB

⇒  70° = ∠EBC + 35° …… [from (iii)]  

⇒ ∠EBC = 70° - 35° = 35° ……. (iv)

From (iii) & (iv), we get

∠ECB = ∠EBC

∴ BE = CE …… [∵ sides opposite to equal angles are equal]

Hence Proved

Answer 2

Answer:

here is the solution

hence it is proved that BE =CE