Answer: The option is D.
Step-by-step explanation:
A line that intersects another line segment and separates it into two equal parts is called a bisector.
In a quadrangle, the line connecting two opposite corners is called a diagonal. We will show that in a parallelogram, each diagonal bisects the other diagonal.
Problem
ABCD is a parallelogram, and AC and BD are its two diagonals. Show that AO = OC and that BO = OD
Strategy
Once again, since we are trying to show line segments are equal, we will use congruent triangles. And here, the triangles practically present themselves. Let’s start with showing that AO is equal in length to OC, by using the two triangles in which AO and OC are sides: ΔAOD and ΔCOB.
There are all sorts of equal angles here that we can use. Several pairs of (equal) vertical angles, and several pairs of alternating angles created by a transversal line intersecting two parallel lines. So finding equal angles is not a problem. But we need at least one side, in addition to the angles, to show congruency.
As we have already proven, the opposite sides of a parallelogram are equal in size, giving us our needed side.
Once we show that ΔAOD and ΔCOB are congruent, we will have the proof needed, not just for AO=OC, but for both diagonals, since BO and OD are alsocorresponding sides of these same congruent triangles.
ABCD is a parallelogram
Given
AD || BC
From the definition of a parallelogram
AD = BC Opposite sides of a parallelogram are equal in size
∠OBC ≅ ∠ODA Alternate Interior Angles Theorem, ∠OCB ≅ ∠OAD Alternate Interior Angles Theorem,
ΔOBC ≅ ΔODA
Angle-Side-Angle
BO=OD Corresponding sides in congruent triangles AO=OC Corresponding sides in congruent triangles.
implying that 
. We then can write:![A-\lambda I = \left [ \begin{array}{cc} 4-\lambda & 2 \\ 5 & 1-\lambda \end{array}\right ]](https://tex.z-dn.net/?f=%20A-%5Clambda%20I%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204-%5Clambda%20%26%202%20%5C%5C%205%20%26%201-%5Clambda%20%5Cend%7Barray%7D%5Cright%20%5D%20)
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