Class 9 – Geometry

Take practice tests in Geometry Midpoint

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  • Mid-point theorem
  • Similarity of triangles
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  • Degree measure of an arc
  • Cyclic quadrilaterals
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If we look at the above figure, we see that three legs of the stool are balanced by a three horizontal rods, these rods are placed parallel to the ground level to balance the person sitting on it, so we can say that the application of midpoint theorem is used in balancing.

Mid-point theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is equal to half of it.
Given A triangle ABC in which D and E are the midpoints of AB and AC respectively.
To prove: DE||BC and .

Construction: Draw CF||BA such that CF meets DE produced at F. Join CF.



Hence proved.

Converse of Midpoint Theorem: A straight line drawn through the midpoint of one side and parallel to another side of a triangle bisects the third side.

Given: ∆ ABC, in which H is the midpoint of and .

To prove: AK = KC

Construction: Draw to meet produced at L.

Proof: In quadrilateral HBCL

Example: In the adjoining figure D and E, are midpoints of AC and BC respectively. Find a) b) ABif DE = 2.1 cm

Solution: In ∆ABC, D and E are the midpoints of AC and BC respectively.
and DE||AB.

(a) (corresponding angles), i.e, =550
(b) AB = 2 × DE = 2 × 2.1 cm = 4.2 cm

Example: Show that the three line segments which join the middle points of the sides of a triangle, divide it into four triangles which are congruent to one another.
Solution: Given in ∆ABC, D,E and F are the midpoints of BC, CA and AB respectively.
To prove: ∆AEF ≅ ∆DEF ≅ ∆FBD ≅ ∆EDC.


Example: ABCD is a trapezium in which. E is the midpoint of BC. EF is drawn parallel to AB which meets AD at F. Prove that 2EF=AB+CD.
Solution: Given ABCD is a trapeziumin which. E is the midpoint of BC. and EF meets AD at F.
To prove: 2EF = AB + CD

Construction: Join BD. BD and EF intersect at G.

AB||CD and EF||AB. (Given) …… (1)

CD||EF (From 1)

In ∆ BCD, E is the mid point of BC and EG||CD

⟹ G is the midpoint BD.( By the converse of the Midpoint Theorem)

In ∆ ADB, G is the midpoint of BD and GF||AB

⟹ F is the midpoint of AD.( By the converse ofthe Midpoint Theorem)

In ∆BCD, the line segment EG joins the midpoint of the sides BC and BD.( By the Midpoint Theorem)

. . . .. (2)

In ∆ ABD, the line segmentGF joins the
midpoints of the sides BD and AD. (By the Midpoint Theorem)
.. . .. .. . .(3)

Adding (2) and (3)

( EG + GF = EF)

2EF = CD + AB.

Hence Proved.

Example: Prove that in a right-angled triangle the median drawn to the hypotenuse is half the hypotenuse in length.
Solution: Given In ∆ ABC, ∠ B = 900. BD is the median drawn to the hypotenuse AC.
To prove:

Construction: Drawn DE||BC such that DEcuts AB at E.

In ∆ABC,
D is the midpoint of AC (Given)

DE||BC (By construction)

∴ E is the midpoint of AB.( By the converse of the Midpoint Theorem) …….(2)

In ∆AED and ∆BED,
AE = EB (From 2)
ED = ED (Common side)
∠AED = ∠BED = 900 (From 3)

∴ ∆AED ≅∆BED.( By SAS criterion of congruency)

AD= BD (c.p.c.t)

(From 1)

Hence Proved.

Example: ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides, AB, BC,CD and DA . AC is a diagonal. Show that:
(i) SRAC and
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution: (i) In ABCD, we have S is the midpoint of AD. R is the midpoint of CD.

and SRAC [by Mid-point theorem ] ——–(A)
(ii),(iii) In △ABC,
P is the midpoint of AB.
Q is the midpoint of BC.

PQ= [by Mid-point theorem]———-(B)
From (A) and (B), we get

⟹PQ = SR and PQSR
∴ PQRS is a parallelogram.

Example: ABCD is a rhombus and P,Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that that quadrilateral PQRS is a rectangle.

Solution: Join AC. In ∆ABC, P and Q are the midpoints of AB and BC.
and PQAC [by Midpoint theorem] ……(A)
In ∆ ADC, R and S are the midpoints of CD and DA.
and SRAC [by Midpoint theorem ] ….(B)
From (A) and (B), we get
=SR and PQACSR ⟹ PQ = SR and PQSR
i.e., One pair of opposite sides of quadrilateral PQRS are equal and parallel.

∴ PQRS is a parallelogram.

Now, in ∆ ERC and ∆EQC,
∠1 = ∠2 [The diagonal of a rhombus bisects the angle at vertex]

CE = CE [Common]
∴ ∆ ERC ≅ ∆EQC [by SAS congruency]

⟹ ∠3= ∠4 [by c.p.c.t]
But ∠3 + ∠4 = 1800 [Linear pair]
⟹ ∠3 = ∠4= 900
∠RQP = 1800 -∠5 [Co-interior angles for PQAC & EQ istransversal]
But ∠5 = ∠3 [Vertically opposite angles]
∴ ∠5 = 900
So, ∠RQP =1800 – ∠5 = 900
∴ One angle of parallelogramPQRS is 900.
Thus, PQRS is a rectangle.

Example: ABCD is a rectangle and P,Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution: In a rectangle ABCD, P is the midpoint of AB, Q is the midpoint of BC, R is the midpoint of CD, S is themidpoint of DA, AC is the diagonal.
Now, in ∆ABC, we have

and PQAC [by Midpoint theorem] ….(A)
Similarly, in ∆ADC, we have
and SRAC [by Midpoint theorem] …(B)
From (A) and (B), we get
PQ = SR and PQ||SR
∴ PQRS is a parallelogram .
Now, in ∆PAS and ∆PBQ, we have
∠A = ∠B [Each 900]
AP = BP [P is mid point of AB]
AS = BQ [AD = BC]
∴ ∆PAS ≅ ∆PBQ [by SAS congruency ]
⟹ PS = PQ [byc.p.c.t]
Also, PS = QR and PQ = SR. [Opposite sides of parallelogram PQRS]
So, PQ = QR = RS = SP
i.e., PQRS is a parallelogram having all of its sides equal.
Hence, PQRS is a rhombus. quadrilateral formed by joining midpoints of sides of a square is a square.

Now that you are well aware of the mid point theorem, the discussion of equal intercepts theorem shall help you in solving problems on the topic of similarity.

Consider the figures given above, AB is a transversal cutting the lines l1and l2 at P and Q respectively. The line segment PQ is calledthe intercept made on the transversal AB by the lines l1 and l2.

Equal intercepts theorem: If a transversal makes equal intercepts on three or more parallel lines then they make equal intercepts on any other transversal intersecting them.
Given: Let be three straight lines such that
Transversal AB makes equal intercepts on l1, l2 andl3, i.e. PQ = QR. Another transversal.

CD makes intercepts KM and MN.
To prove: KM = MN
Construction: Join PN which cuts the L2 at O.

Proof: PQ = QR and QOl3(Given)

In ∆PRN, O is the midpoint of PR (PQ = QR)
PO = ON (By converse of Midpoint Theorem)

O is the mid point of PN,

M is the midpoint of NK, i.e., KM = MN (by converse of Midpoint theorem)

Hence Proved.

In D and E are midpoints on AB, AC and G, H are midpoints on AD, AE respectively then find

i) The relation between GH and BC. ii) GH divided AB and AC in the ratio of:


i) In D and E are midpoints on AB and AC
By mid point theorem

In G and H are midpoints on AD and AE
By mid point theorem

From(i) and (ii)

ii) From (i) and (ii)