Chapter 7.2, Problem 18PSC

Solution

To Prove: ABCD is a rectangle.

Given information: $\overline{EF}$ is the median to $\overline{AC}$ , $\angle CBD\cong \angle ADB$ , $\overline{CD}$ is the base of isosceles $\Delta FDC$ is provided in the question which will help to find the solution.

Proof: Here, EF is median to AC

So that, AF=FC

And, $\angle CBD=\angle ADB$ (given)

Hence, DC=AB (sides opposite to equal angles are equal)

And, $\Delta FDC$ is isosceles.

So, FD=FC

AF=FC

So that, AF=FD=FC

So that, here diagonals bisect at same point.

Hence, AD=BC and AB=CD

So that, ABCD is a rectangle.