Suppose ABC be a Triangle where Ang.A=Ang.B and D&E are the points on ur figure..Then Clearly AC=BCNOW,we have AD=BEso,AC-AD=BC-BEI.E.CD=CEOR,CD/CE=1ALSO,AC/BC=1SO,CD/CE=AC/BCOR,CD/AC=CE/BCNOW,IN TRIANGLE.CDE AND TRIANGLE.CABCD/AC=CE/BCALSO,ANG.DCE=ANG.ACB (COMMON)SO.BOTH TRIANGLE ARE SIMILAR SO,BY B.P.T...DE PARALLEL AB..

Proof:

Given: In triangle ABC, angle A = angle B and AD = BE.

To prove: DE || AB.

Proof:

Step 1: Draw a line segment CF parallel to DE, intersecting AB at F.

Step 2: Now, we need to prove that DE || AB. To prove this, we will prove that angles ADE and ACF are equal.

Step 3: In triangle ABD, angle A = angle B (Given).

Step 4: In triangle ADE, angle A = angle DAE (Angles opposite to equal sides are equal).

Step 5: In triangle ACF, angle A = angle CAF (Angles opposite to equal sides are equal).

Step 6: From Step 4 and Step 5, we have angle DAE = angle CAF.

Step 7: In triangle ADE, angle DAE = angle ADE (Angle sum property of a triangle).

Step 8: From Step 6 and Step 7, we have angle ADE = angle CAF.

Step 9: From Step 8, we can conclude that angles ADE and ACF are equal.

Step 10: Now, since angles ADE and ACF are equal, it implies that DE is parallel to CF (Alternate Interior Angles Theorem).

Step 11: But CF is parallel to DE, so we can conclude that DE is parallel to AB (Transitive Property of Parallel Lines).

Step 12: Hence, we have proved that DE || AB.

Therefore, in triangle ABC, if angle A = angle B and AD = BE, then DE is parallel to AB.