It is known that in triangle ABC, Be and CF are bisectors of angles, and d is the midpoint of EF. If the distances from D to the three sides of the triangle BC, AB and AC are X, Y, z Y and Z respectively, it is proved that X = Y+Z.

Prove; The point passing through E is the high intersection of AB and BC, and BC is at M and N respectively.

The intersection f is the intersection of p and q on AC and BC, respectively.

According to the distance between the point on the bisector of the angle and both sides of the angle is equal, we can know that FQ = FP and EM = en.

Make a high intersection between point D on BC and point O on BC.

Intersection D is the intersection of AB at point H, and intersection D is the intersection of AB at point J. 。

Then x = do, y = hy, z = DJ.

Because d

Is the midpoint, angle ANE= angle AHD=90 degrees. So HD is parallel to me, ME=2HD.

It can also be proved that FP = 2dj.

Because FQ=FP, EM=EN.

FQ=2DJ, um =2HD.

And because the angles FQC, DOC and ENC are all 90 degrees, the quadrilateral FQNE is a right trapezoid, and d is the midpoint, so 2do = FQ+en.

because

FQ=2DJ, um =2HD .So do = HD+JD.

Because x = do, y = hy and z = DJ, x = y+z.

This handwriting is really difficult to type. You must understand it. When was this the geometry problem for the senior high school entrance examination? Look carefully.

If you don't understand, it must be correct. I don't want to work in vain. 20 points is really tiring. You must understand it. If not, ask me again. oh