Just a trick on the Triangle

Drop perpendiculars from D and E on AB & AC respectively.Call the foot of these perpendiculars as M and N .Let BD,DE,EC be 3x,5x,4x respectively.

Now,EN=NC.Find each one of them in the form of x using the pythagoras theorem.Now use pythagoras theorem in triangle ABC to find AB & AC in the form of x.Subtract NC from AC to get AN.Now, in triangle AEN, find angle EAN using trigonometry.

Similarly, find angle DAM.Add the two findings and subtract the sum from 90,you will get 45 as the answer...........Please send a better solution if possible as my solution would require trigonometric tables to reach to the answer

There's a little trick at first. Rotating the configuration about A by 90, the point B goes to the point C. Let P denote the image of the point D under this rotation. Then CP = BD and ∠ACP = ∠ABC = 45◦, so ECP is a right-angled triangle with CE : CP = 4 : 3. Hence PE = ED. It follows that ADEP is a kite with AP = AD and PE = ED. Therefore AE is the angular bisector of ∠PAD. This implies that ∠DAE = ∠PAD/2 = 45.