Question

In a triangle ABC angle BCA=90. Points E and F lie on the hypotenuse AB such that AE=AC and BF=BC Find angle ECF

Answer 1

Given, ∠C = 90°As angles opposite to equal sides of a ∆ are equal, so∠1 =∠2   ( Since, BF = BC)∠3 =∠4   ( Since, AE = AC)Now, ∠ECF = ∠2+∠4 -∠C..........(i)Now, in ∆ECF∠1 +∠3 + ∠ECF = 180°⇒∠1 +∠3 +-∠C =180° (from (i))⇒2∠2+2 ∠4  = 180° +90°⇒∠2+∠4 =135°.......(ii)∠ECF = 135°-90°=45°

Answer 2

Answer:

45°

Step-by-step explanation:

From , AE = AC and BC = BF, we have

⇒∠AEC = 1/2 (180° – ∠A) = 90° – 1/2 ∠A,

⇒∠BFC= 1/2 (180° – ∠B) = 90° – 1/2 ∠B,

Therefore

⇒∠ECF = 180° – ∠AEC – ∠BFC

=  1/2( ∠A+∠B) = 45°