In Triangle ABC, medians BE and CD are produced respectively to points X and Y such that CD=DX and BE=EY as shown in the figure. Show that the points X, A and Y are collinear. Also show that A is the mid point of XY

To show that points X, A, and Y are collinear, we can use the concept of transversals.

Step 1: Start by drawing a diagram of Triangle ABC with medians BE and CD produced to points X and Y, respectively.

Step 2: Assume that X, A, and Y are collinear. This means that there exists a straight line that passes through all three points.

Step 3: From the given information, we know that CD = DX and BE = EY. This implies that triangle BDX and triangle BEY are congruent (by side-side-side congruence).

Step 4: By triangle congruence, we can conclude that angle BDX = angle BEY.

Step 5: Recall that angles on the same side of a transversal line and between parallel lines are supplementary.

Step 6: From above, angle BDX = angle BEY = angle DCB (since medians BE and CD are produced from the sides of the triangle).

Step 7: Since angles BDX and DCB are equal, and angles BEY and DCB are equal, it follows that angles BDX and BEY are equal.

Step 8: By the transitive property of equality, we can conclude that angle BDX = angle BEY = angle BDA.

Step 9: From step 4 and step 8, we can conclude that angle BDA is common to both triangle BDX and triangle BEY.

Step 10: Since angle BDA is common to both triangles, and angles BDA and DBX have the same measure, and angles BDA and BEY have the same measure, it follows that angle DBX = angle DBA = angle DBY.

Step 11: By the converse of the Angle Bisector Theorem, we can conclude that lines AX and AY bisect angles BDX and BEY, respectively.

Step 12: Since angles DBX and DBY are adjacent angles on a straight line, and they are each bisected by line segment AX and AY, we can conclude that lines AX and AY are the same line.

Step 13: Therefore, points X, A, and Y are collinear.

To show that A is the midpoint of XY, we can use the concept of midpoints.

Step 14: Since points X, A, and Y are collinear, we can consider line segment XY.

Step 15: From the given information, we know that BE = EY. Since BE is a median, it divides side BC into two equal parts.

Step 16: Therefore, we can conclude that line segment BA is equal to line segment AY.

Step 17: Similarly, from the given information that CD = DX, and CD is a median, it divides side AC into two equal parts.

Step 18: Therefore, we can conclude that line segment AX is equal to line segment BA.

Step 19: Combining step 16 and step 18, we can conclude that line segment AX is equal to line segment AY.

Step 20: Since line segment AX is equal to line segment AY, A is the midpoint of XY.

Therefore, we have shown that points X, A, and Y are collinear, and A is the midpoint of XY.

To show that the points X, A, and Y are collinear, we can use the concept of similar triangles and the properties of medians.

1. Let's first show that ∆ABX and ∆ACY are similar triangles. Both triangles share ∠A and have one pair of sides in proportion: AX and AY are both medians of ∆ABC, so their lengths are equal.

2. Since AX = AY, and BE = EY, it follows that AX = AY = EY.

3. Now, let's consider ∆ABX and ∆ADC. These two triangles have ∠A in common and two pairs of proportional sides: AB || DC (medians), and AX = AY = EY.

4. By using the corresponding sides of ∆ABX and ∆ADC, we can conclude that ∆ABX ∼ ∆ADC by the Side-Angle-Side (SAS) criterion for similarity.

5. Therefore, the corresponding angles of ∆ABX and ∆ADC are equal: ∠XAB = ∠DAC.

6. Since ∠XAB = ∠DAC and ∠BAC = ∠DAC (since BAC is shared between ∆ABC, ∆ABX, and ∆ADC), we have ∠XAB = ∠BAC.

7. This shows that point X lies on the line passing through points A and C.

8. By a similar argument, we can prove that point Y also lies on the line passing through points A and C.

9. Therefore, the points X, A, and Y are collinear.

To show that A is the midpoint of XY:

10. Since AX = AY (as proven earlier), it means that AX = AY = 1/2 XY.

11. This indicates that A is the midpoint of XY: it divides XY into two equal halves.

Consider triangles AEY and CEB.

AE=CE (BE bisects AC)
EY=EB (median is extended its own length)
angles AEY and CEB are congruent (vertical angles)

So, the two triangles are congruent (SAS). That means that AY=BC and is parallel.

Now do the same with triangles ADX and BDC.

The rest follows.