Let's break down this geometry problem step by step.

Given:

Triangle ABC is right-angled at A (∠BAC=90∘).
AB<AC.
BD is the angle bisector of ∠ABC, with D on AC.
DM⊥BC at M.
Part 1: Prove AD<AB

Angle Bisector Theorem: Since BD bisects ∠ABC, by the Angle Bisector Theorem, we have:

DCAD​=BCAB​
Hypotenuse is the longest side: In right-angled triangle ABC, BC is the hypotenuse, so BC>AB.
Comparing ratios: Since BC>AB, we have BCAB​<1.

Therefore, DCAD​<1, which implies AD<DC.
Consider triangle ABD: In triangle ABD, we have angles ∠BAD=90∘ and ∠ABD=21​∠ABC. Since ∠ABC is an angle in a right-angled triangle, ∠ABC<90∘, so ∠ABD<45∘.
Angle sum of triangle ABD: ∠ADB=180∘−∠BAD−∠ABD=180∘−90∘−∠ABD=90∘−∠ABD.

Since ∠ABD<45∘, we have ∠ADB>90∘−45∘=45∘.
Comparing angles in triangle ABD: We have ∠ABD<45∘ and ∠ADB>45∘, so ∠ABD<∠ADB.
Relationship between sides and opposite angles: In triangle ABD, the side opposite the larger angle is longer. Therefore, AD<AB (since AD is opposite ∠ABD and AB is opposite ∠ADB).
Part 2: Prove AK, EI, MD are concurrent

New Given Information:

From M, a line perpendicular to AB intersects AC at E (ME⊥AB).
K is the midpoint of ME.
I is the intersection point of BD and AM.
We need to prove that AK, EI, and MD are concurrent (they intersect at a single point).

Let's use vector methods or coordinate geometry to solve this concurrency problem.

Using Coordinate Geometry:

Place the origin at A (0, 0).
Let B be on the x-axis at (b,0) where b>0.
Let C be on the y-axis at (0,c) where c>b>0 (since AC>AB).
Equation of line BC: The slope of BC is 0−bc−0​=−bc​. The equation of BC is y−0=−bc​(x−b), which simplifies to bx+cy=bc.
Coordinates of D (on AC, y-axis): Let D=(0,d). Since D lies on the angle bisector of ∠ABC, the distances from D to AB (x-axis) and BC are equal.

Distance from D to AB = ∣d∣.

Distance from D to BC = b2+c2​∣b(0)+c(d)−bc∣​=b2+c2​∣cd−bc∣​=b2+c2​∣c(d−b)∣​.

∣d∣=b2+c2​∣c(d−b)∣​.

Since D is on AC, 0<d<c. Also, since AB<AC, ∠C<∠B, so D will be closer to A.
Coordinates of M (DM⊥BC): The line DM is perpendicular to bx+cy=bc, so its slope is cb​. The equation of DM is y−d=cb​(x−0), which is cy−cd=bx. M is the intersection of bx+cy=bc and bx−cy=−cd.

Adding the two equations: 2bx=bc−cd⟹xM​=2bc(b−d)​.

Substituting xM​ into bx+cy=bc: b2bc(b−d)​+cy=bc⟹2c(b−d)​+cy=bc⟹cy=bc−2bc−cd​=22bc−bc+cd​=2bc+cd​.

yM​=2cc(b+d)​=2b+d​.

So, M=(2bc(b−d)​,2b+d​).
Coordinates of E (ME⊥AB): Since ME⊥AB (x-axis), the x-coordinate of E is the same as M, and E lies on AC (y-axis). So, E=(0,2b+d​).
Coordinates of K (midpoint of ME): K=(20+2bc(b−d)​​,22b+d​+2b+d​​)=(4bc(b−d)​,2b+d​).
Coordinates of I (intersection of BD and AM):

Equation of BD (line passing through B(b,0) and D(0,d)): bx​+dy​=1⟹dx+by=bd.

Equation of AM (line passing through A(0,0) and M(2bc(b−d)​,2b+d​)): y=2bc(b−d)​2b+d​​x=c(b−d)b(b+d)​x.

Intersection I: dx+b(c(b−d)b(b+d)​x)=bd.

xI​(d+c(b−d)b2(b+d)​)=bd.

xI​(c(b−d)cd(b−d)+b2(b+d)​)=bd.

xI​=cd(b−d)+b2(b+d)bcd(b−d)​.

yI​=c(b−d)b(b+d)​⋅cd(b−d)+b2(b+d)bcd(b−d)​=cd(b−d)+b2(b+d)b2d(b+d)​.

So, I=(cd(b−d)+b2(b+d)bcd(b−d)​,cd(b−d)+b2(b+d)b2d(b+d)​).
Coordinates of M: M=(2bc(b−d)​,2b+d​).
Equation of AK (line passing through A(0,0) and K(4bc(b−d)​,2b+d​)): y=4bc(b−d)​2b+d​​x=c(b−d)2b(b+d)​x.
Equation of EI (line passing through E(0,2b+d​) and I(cd(b−d)+b2(b+d)bcd(b−d)​,cd(b−d)+b2(b+d)b2d(b+d)​)):

Slope of EI = cd(b−d)+b2(b+d)bcd(b−d)​−0cd(b−d)+b2(b+d)b2d(b+d)​−2b+d​​.

Equation: y−2b+d​=slope(x−0).
Equation of MD (line passing through M(2bc(b−d)​,2b+d​) and D(0,d)):

Slope of MD = 2bc(b−d)​−02b+d​−d​=c(b−d)/bb−d​=c(b−d)b(b−d)​=cb​.

Equation: y−d=cb​(x−0)⟹cy−cd=bx.
To prove concurrency, we need to show that there exists a point (x,y) that satisfies the equations of AK, EI, and MD simultaneously. This algebraic approach is quite complex.

Using a Geometric Approach (Ceva's Theorem or Menelaus' Theorem might be applicable, but the setup isn't directly fitting a triangle with points on its sides):

Consider using mass point geometry or a clever construction.

Let's consider the properties of the points and lines involved.

BD is the angle bisector.
M is such that DM is perpendicular to BC.
E has the same x-coordinate as M and lies on AC.
K is the midpoint of ME.
I is the intersection of the angle bisector BD and AM.
Consider triangle AME. K is the midpoint of ME. AK is a median.

Consider triangle ABM. BD intersects AM at I.

Let's think about the implications of ME⊥AB. Since AB is along the x-axis, ME is vertical.

Consider the coordinates again. The x-coordinate of M and E is the same. E lies on AC (y-axis), so the x-coordinate of E is 0. This implies 2bc(b−d)​=0, so b=d (since c=0,b=0).

If b=d, then D has coordinates (0,b).

Since D is on the angle bisector, the distance to AB and BC are equal:

∣b∣=b2+c2​∣c(b−b)∣​=0, which implies b=0, a contradiction.

There must be a mistake in the coordinate geometry setup or interpretation.

Let's rethink the geometry without coordinates.

Consider triangle ABE. MK is parallel to AB (since ME⊥AB and the perpendicular from M to AB is along ME). K is the midpoint of ME.

Consider the properties of point I, the intersection of the angle bisector BD and AM.

Consider triangle BCM and the line segment MD.

Let's use a property related to the symmedian.

Consider the case where triangle ABC is isosceles right-angled at A. Then the angle bisector BD makes a 45-degree angle with AB and BC.

Let's use a different approach focusing on ratios and Menelaus' Theorem.

Consider triangle AME and the line BKI.

Consider triangle ABM and the line D-I-C.

This problem seems more involved than initially apparent and likely requires a specific geometric insight or theorem application. The coordinate geometry approach became cumbersome due to the complexity of finding the coordinates of the intersection point I.

Without a clear geometric theorem that directly applies to this configuration, proving the concurrency of AK, EI, and MD is challenging within this response format. A diagram drawn accurately might provide visual clues.

Final Answer: The final answer is AD<AB​