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Ex.8.2 Q7 Quadrilaterals Solution - NCERT Maths Class 9

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Question

\(ABC\) is a triangle right angled at \(C\). \(A\) line through the mid-point \(M\) of hypotenuse \(AB\) and parallel to \(BC\) intersects \(AC\) at \(D\). Show that

(i) \(D\) is the mid-point of \(AC\)

(ii) \( \rm{MD} \perp \rm{AC} \)

(iii)  \(\begin{align} \rm{CM}=\rm{MA}=\frac{1}{2} \rm{AB} \end{align}\)

 Video Solution
Quadrilaterals
Ex 8.2 | Question 7

Text Solution

What is known?

\(ABC\) is a triangle right angled at \(C\). A line through the mid-point M of hypotenuse \(AB\) and parallel to \(BC\) intersects \(AC\) at \(D.\)

What is unknown?

How we can show that

(i) \(D\) is the mid-point of \(AC\)

(ii) \({MD }\!\!~\!\!\bot {AC}\)

(iii) \({CM}={MA}=\frac{1}{2}{ AB}\)

Reasoning:

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side. To show \(MD\) perpendicular to \(AC\) we know that sum of interior angles is \(180\). Also to get \(CM=MA= \)half of AB we can show two triangles congruent

Steps:

(i) In \(\Delta {ABC,}\)

It is given that \(M\) is the mid-point of \(AB\) and \(MD\) \(\parallel \) \(BC\).

Therefore, \(D\) is the mid-point of \(AC\). (Converse of mid-point theorem)

(ii) As \(DM\) \(\parallel \) \(CB\) and \(AC\) is a transversal line for them, therefore,

\[\begin{align} &\angle {MDC}+\angle {DCB}=180^{\circ}\\&({Co} \text { -interior angles })\\ &\angle {MDC}+90^{\circ}=180^{\circ}\\ &\angle {MDC}=90^{\circ}\\ &\therefore {MD} \perp {AC}\end{align}\]

(iii) Join \(MC\).

In \(\Delta {AMD}\) and \(\Delta {CMD}\)

\(AD = CD\) (D is the mid-point of side \(AC\))

\(\begin{align}  \angle ADM= \angle CDM \; (Each\; 90^0 ) \end{align}\)

\(DM = DM\) (Common)

\( \begin{align} \therefore \Delta {AMD}\cong \Delta {CMD}\end{align}\) (By SAS congruence rule)

Therefore, \(AM  = CM\) (By CPCT)

However, \( \begin{align}  AM = \frac{1}{2} AB \end{align}\) (M is the mid-point of AB)

Therefore, it can be said that

\[ \begin{align} CM= AM = \frac{1}{2} AB \end{align}\]

 Video Solution
Quadrilaterals
Ex 8.2 | Question 7
  
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