Question

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

Answer 1

Answer:

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Answer 2

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°

∴ ∠ECB = 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.

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