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1 answer:
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<u>Answer-</u>
<em>The correct answer is</em>
<em>∠BDC and ∠AED are right angles</em>
<u>Solution-</u>
In the ΔCEA and ΔCDB,
As this common to both of the triangle.
If ∠BDC and ∠AED are right angles, then
Now as
∠BCD = ∠ACE and ∠BDC = ∠AED,
∠DBC and ∠EAC will be same. (as sum of 3 angles in a triangle is 180°)
Then, ΔCEA ≈ ΔCDB
Therefore, additional information can be used to prove ΔCEA ≈ ΔCDB is ∠BDC and ∠AED are right angles.
Answer:
x=πn, n∈Z
Step-by-step explanation:
if sin²2x=1-cos²2x, then
1-cos²2x-2cos2x+2=0; ⇒ cos²2x+2cos2x-3=0; ⇔ (cos2x+3)(cos2x-1)=0;