In isosceles △ABC with AC = AB, points D ∈ AB and E ∈ AC so that DB = EC. Point P is the intersection point of DC and EB, m∠ADC = 107°, m∠ABE = 45°. Find m∠BAP.
Please show work.

9514 1404 393

  14°

Step-by-step explanation:

∠ADP is an external angle to ΔDBP, so ...

  ∠ADP = ∠DBP + ∠DPB

  107° = 45° + ∠DPB

  ∠DPB = 107° -45° = 62°

∠DPB is an external angle to isosceles triangle PBC, so ...

  ∠DPB = 2×∠PBC

  62° = 2×∠PBC

  ∠PBC = 62°/2 = 31°

Of course, the angle sum theorem applies at vertex B, so ...

  ∠ABC = ∠ABP +∠PBC

  ∠ABC = 45° +31° = 76°

AP is the angle bisector of angle A, so, if extended to BC would be perpendicular to BC. Then ∠BAP is complementary to ∠ABC:

  ∠BAP = 90° -∠ABC

  ∠BAP = 90° -76°

  ∠BAP = 14°

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

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step-by-step explanation: