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