## Selina Concise Mathematics Class 9 ICSE Solutions Mid-point and Its Converse [ Including Intercept Theorem]

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APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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**Selina ICSE Solutions for Class 9 Maths Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]**

**Exercise 12(A)**

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**Exercise 12(B)**

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**Solution 12:**

Given ABCD is parallelogram, so AD = BC, AB = CD.

Consider triangle APB, given EC is parallel to AP and E is midpoint of side AB. So by midpoint theorem, C has to be the midpoint of BP.

So BP = 2BC, but BC = AD as ABCD is a parallelogram.

Hence BP = 2AD

Consider triangle APB, AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.

Hence Proved

**Solution 13:**

Consider trapezium ABCD.

Given E and F are midpoints on sides AD and BC, respectively.

Consider LHS,

AB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG

So AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF

AB + CD = 2EF

Hence Proved

**Solution 14:**

Given Δ ABC

AD is the median. So D is the midpoint of side BC.

Given DE || AB. By the midpoint theorem, E has to be midpoint of AC.

So line joining the vertex and midpoint of the opposite side is always known as median. So BE is also median of Δ ABC.

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