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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 