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Class 9 FRANK Solutions Maths Chapter 15 - Mid-point and Intercept Theorems

Practise Frank Solutions for ICSE Class 9 Mathematics Chapter 15 Mid-point and intercept theorems at TopperLearning. If you find questions on working with the mid-point of sides of a triangle mind-boggling, our reference solutions will help you. Whether it is sides of a rhombus or a parallelogram, you can check the steps that experts will use to solve Maths problems.

Revise textbook questions on the Mid-point theorem and Intercept theorem effectively with Frank textbook solutions. For more ICSE Class 9 Maths resources, explore our Selina solutions, online practice tests, videos and more.

Mid-point and Intercept Theorems Exercise Ex. 15.1

Solution 3(a)


Solution 3(b)


Solution 3(c)

Solution 5


Solution 11


Solution 15(a)

Let us draw a diagonal AC which meets PQ at O as shown below:





Solution 18

Construction: Draw DS BF, meeting AC at S.


Solution 24


Solution 1

Solution 2

Solution 4

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 12

Solution 13

Solution 14

Solution 16

Solution 17

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Mid-point and Intercept Theorems Exercise Ex. 15.2

Solution 1




Solution 2

Solution 3

Solution 4

Solution 5

The figure is as below:

Let ABCD be a quadrilateral where P, Q, R, S are the midpoints of sides AB, BC, CD, DA respectively.

Diagonals AC and BD intersect at right angles at point O. We need to show that PQRS is a rectangle

Solution 6

Note: This question is incomplete.

According to the information given in the question,

F could be any point on BC as shown below:

So, this makes it impossible to prove that DP = DE, since P too would shift as F shift because P too would be any point on DE as F is.

Note: If we are given F to be the mid-point of BC, the result can be proved.

Solution 7


From the figure EF AB and E is the midpoint of BC. 

Therefore, F is the midpoint of AC. 

Here EF BD, EF = BD as D is the midpoint of AB. 

BE DF, BE = DF as E is the midpoint of BC. 

Therefore BEFD is a parallelogram.

Remark: Figure modified

Solution 8

Solution 9

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