NCERT Solutions for Class 9 Maths Chapter 8
Important NCERT Questions
Quadrilaterals
NCERT Books for Session 2022-2023
CBSE Board and UP Board Others state Board
EXERCISE 8.2
Page No:150
Questions No:7
(i) D is the mid-point of AC
(ii) MD⊥ AC
(iii) CM = MA =1 AB/2
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(i) In ΔABC,
M is mid-point to AB [∵ Given]
and DM ∥ BC [∵ Given]
Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem]
(ii) ∠ADM = ∠ACB [∵ Corresponding angles]
⇒ ∠ADM = 90° [∵ ∠ACB 90°]
Hence, MD ⊥ AC
(iii) In ΔAMD = ΔCMD,
AD = DC [∵ Proved above]
∠ADM = ∠CDM [∵ Each 90°]
DM = DM [∵ Common]
Hence, ΔAMD ΔCMD [∵ SAS Congruency rule]
AM = CM [∵ CPCT]
But AM = (1/2) AB [∵ Given]
Therefore, CM = AM = (1/2) AB