# FRANK Solutions for Class 9 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.

## Chapter 15 - Mid-point and Intercept Theorems Exercise Ex. 15.1

InABC, D, E, F are the midpoints of BC, CA and AB respectively. Find FE, if BC = 14 cm

InABC, D, E, F are the midpoints of BC, CA and AB respectively. Find DE, if AB = 8 cm

InABC, D, E, F are the midpoints of BC, CA and AB respectively. Find

FDB ifACB = 115°.

InABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that:

a) QAP is a straight line.

b) A is the mid-point of PQ

In a right-angled triangle ABC. ABC = 90° and D is the midpoint of AC. Prove that BD =AC.

In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find:

a) PQ, if AB = 12 cm and DC = 10 cm.

b) AB, if DC = 8 cm and PQ = 9.5 cm

c)
DC, if AB = 20 cm and PQ = 14 cm** **

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

AD is a median of side BC ofABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF : AC = 1 : 3.

Construction: Draw DS ∥ BF, meeting AC at S.

InABC, D, E and F are the midpoints of AB, BC and AC.

a. Show that AE and DF bisect each other.

b. If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.

## Chapter 15 - Mid-point and Intercept Theorems Exercise Ex. 15.2

In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that:

a. GEA GFD

b. HEB HFC

c. EGFH is a parallelogram

InABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that:

a. Q A and P are collinear.

b. A is the mid-point of PQ

In AABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC.

In the given figure, the lines l, m and n are parallel to each other. D is the midpoint of CE. Find:

a. BC, b. EF, c. CG and d. BD.

The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the midpoints of quadrilateral ABCD is a rectangle.

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

InABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.

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.

InABC, D and E are the midpoints of the sides AB and BC respectively. F is any point on the side AC. Also, EF is parallel to AB. Prove that BFED is a parallelogram.

Remark: Figure is incorrect in Question

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

In the given figure, PS = 2RS. M is the midpoint of QR. If TR ∥ MN ∥ QP, then prove that:

In the given figure, T is the midpoint of QR. Side PR ofPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR =PQ.

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