In ∆ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.
Step-by-step solution
To find: Prove that MN bisects AD
Idea: MN is parallel to BC, so inside the smaller triangle ABD it is a line through the midpoint of one side parallel to another side. Such a line bisects the third side, AD.
- In ∆ABC, M and N are midpoints of AB and AC, so MN ‖ BC (Midpoint Theorem).1 mark
- Let MN meet AD at E. D lies on BC, so ME ‖ BD.½ mark
- In ∆ABD, M is the midpoint of AB and ME ‖ BD, so E is the midpoint of AD (converse of the Midpoint Theorem). Hence AE = ED: MN bisects AD.1½ marks
Check: A(0, 6), B(−4, 0), C(8, 0), D(2, 0): M(−2, 3), N(4, 3), so MN is the line y = 3. Midpoint of AD = (1, 3), which lies on y = 3 ✓.
Answer to write in the exam
MN ‖ BC (Midpoint Theorem in ∆ABC)
Let MN meet AD at E ⇒ ME ‖ BD
In ∆ABD: M midpoint of AB, ME ‖ BD ⇒ E midpoint of AD (converse of Midpoint Theorem)
∴ AE = ED, i.e. MN bisects AD.
Common mistakes that cost marks
- Applying the converse in ∆ABC instead of ∆ABD. The bisected side must be AD, so use the triangle that has AD as a side.
- Assuming D is the midpoint of BC. The result holds for any point D on BC.
- Writing “MN = ½AD”. MN bisects AD; it is not half of it.
How this can come in the exam
In ∆PQR, S and T are the midpoints of PQ and PR. U is a point on QR with PU = 9 cm. ST meets PU at V. Find PV.
Show answer
ST ‖ QR (Midpoint Theorem), so in ∆PQU the line through the midpoint S parallel to QU bisects PU (1 mark). PV = 9 ÷ 2 = 4.5 cm (1 mark).Try one yourself
In ∆ABC, M and N are the midpoints of AB and AC, and AD is the altitude to BC. Show that MN bisects AD.
Show answer
D is a point on BC, so by the same argument (MN ‖ BC, then the converse of the Midpoint Theorem in ∆ABD) MN meets AD at its midpoint.
More questions like this
- In a quadrilateral ABCD, suppose AB ‖ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ‖ AB. (Why did we assume AB ≠ CD?)
- Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
- Suppose PQRS is the Varignon parallelogram of ABCD.
- Suppose we have a tiling of the plane. Consider any vertex. As we go around this vertex and consider the angles made by consecutive lines, the total of these angles must be 360°. This suggests an idea. What if we take 4 copies of SOME and fit them together around a common point so that each angle is used once as we go around?
- There are multiple ways of doing this, as shown in the figure. Can you use any of these ways to continue fitting further copies of SOME to tile the plane? Try it with the 15 copies you made!
All Quadrilaterals and parallelograms questions · All maths questions