MathDB

In triangle

ABC,

we have

\angle A \leq 90^\circ

and

\angle B = 2 \angle C.

The interior bisector of the angle

C

meets the median

AM

D.

Prove that

\angle MDC \leq 45^\circ.

When does equality hold?
[asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqqqzz = rgb(0,0,0.6); pen ffqqtt = rgb(1,0,0.2); pen qqwuqq = rgb(0,0.39,0); draw((3.14,5.81)--(3.23,0.74),ffqqtt+linewidth(2pt)); draw((3.23,0.74)--(9.73,1.32),ffqqtt+linewidth(2pt)); draw((9.73,1.32)--(3.14,5.81),ffqqtt+linewidth(2pt)); draw((9.73,1.32)--(3.19,2.88)); draw((3.14,5.81)--(6.71,1.05)); pair parametricplot0_cus(real t){ return (0.7*cos(t)+5.8,0.7*sin(t)+2.26); } draw(graph(parametricplot0_cus,-0.9270442638657642,-0.23350086562377703)--(5.8,2.26)--cycle,qqwuqq); dot((3.14,5.81),ds); label("

A

", (2.93,6.09),NE*lsf); dot((3.23,0.74),ds); label("

B

", (2.88,0.34),NE*lsf); dot((9.73,1.32),ds); label("

C

", (10.11,1.04),NE*lsf); dot((3.19,2.88),ds); label("

X

", (2.77,2.94),NE*lsf); dot((6.71,1.05),ds); label("

M

", (6.75,0.5),NE*lsf); dot((5.8,2.26),ds); label("

D

", (5.89,2.4),NE*lsf); label("

\alpha

", (6.52,1.65),NE*lsf); clip((-3.26,-11.86)--(-3.26,8.36)--(20.76,8.36)--(20.76,-11.86)--cycle); [/asy]

Problem Statement