Prove: Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects. (PS: Please do not use “short-cut or the magic formula — Pythagoras’s theorem; if you do, you are requested to prove Pythagoras’s theorem using plane geometry!) (PS:2: Remember Euclid’s geometry builds up from “scratch”…a small step at a time….axioms, definitions, lemmas, …; you can only use the theorems used and proved so far in “class”).
If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the corresponding sides of the other; then the base of that which has the greater angle is greater than the base of the other.
Prove converse of the above: If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.
(i) The perpendicular is the shortest line that can be drawn to a given straight line from a given point. (ii) Obliques which make equal angles with the perpendicular are equal. (iii) Of two obliques the less is that which makes the smaller angle with the perpendicular. Prove all these.
If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides are either equal or supplementary, and in the former case, the triangles are equal in all respects.