The vertices of a triangle are A (-1,-4), B (2,-4) and C (2,-1). Use the distance formula and Pythagoras' theorem to show that triangle ABC is an isosceles right-angled triangle

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

List the sides of δrst in in ascending order (shortest to longest) if: a ∠r =x+28°, m∠s = 2x+16°, and m∠t = x+12°

A3-dimensional figure has a square base and 4 lateral faces that meet at a point. explain the difference between the cross sections when a slice is made perpendicular to the base through the vertex, and when a slice is made perpendicular to the base and not through the vertex