Given: YX is parallel to BC, AX=AY, BV=CU
Prove: AB=AC

Proof:  By similar triangles AUY and BUC, AU/BU = AY/BC.  By similar triangles AVX and CVB, AX/BC = AV/CV.  And since AX=AY, we have

AU/BU = AY/BC = AX/BC = AV/CV,

By side-angle-side, triangles AUV and ABC are similar, so UV is parallel to BC, which establishes that VUBC is a trapezoid.  And it's a trapezoid with congruent diagonals, we're given.   A trapezoid with congruent diagonals is isosceles (lemma 2, below).  Since trapezoid VUBC is isosceles, so is triangle ABC, and so AB = AC, and the result is proved. 

Lemma 1: An isosceles trapezoid has congruent diagonals.

By definition, an isosceles trapezoid is a trapezoid with equal base angles, and therefore by the Pythagorean Theorem equal left and right sides.  The diagonals of an isosceles trapezoid are congruent because they form congruent triangles with the other two sides of the trapezoid, which is shown using side-angle-side.

Lemma 2: A trapezoid with congruent diagonals is isosceles.

Using the diagram, let VUBC be a trapezoid, with bases VU and BC, and congruent diagonals BV and CU.  Now, triangles BCU and CBV have the same height and the same base, so they have the same area.  The area of a triangle is the half the product of a pair of sides times the sine of the included angle, so sine of BCU equals the sine of CBV, and as both angles are acute (or both obtuse) they are equal.  Now, by side-angle-side, triangles BCU and CBV are congruent, and so VUBC is isosceles.