# Ex.6.3 Q14 Triangles Solution - NCERT Maths Class 10

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## Question

Sides $$AB$$ and $$AC$$ and median $$AD$$ of a $$\triangle ABC$$ are respectively proportional to sides $$PQ$$ and $$PR$$ and median $$PM$$ of another $$\triangle PQR.$$

Show that $$\Delta ABC\text{ }\sim{\ }\text{ }\Delta PQR$$ .

Diagram

## Text Solution

Reasoning:

As we know if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

This criterion is referred to as the $$\rm SAS$$ (Side–Angle–Side) similarity criterion for two triangles.

Steps:

Produce $$AD$$  to $$E$$ so that $$AD = DE$$ . Join $$CE$$ Similarly, produce $$PM$$ to $$N$$ such that $$PM = MN$$, and Join $$RN$$.

In $$\Delta ABD$$  and $$\Delta CDE$$

$$AD = DE$$ [ By Construction]
$$BD = DC$$ [AP is the median]

$$\angle ADB = \angle CDE$$ [Vertically opposite angles]
Therefore,
$$\Delta ABD \cong \Delta CDE$$ [By SAS criterion of congruence
$$\Rightarrow AB = CE$$ [ CPCT] ... (i)

Also, in $$\Delta PQM$$ and $$\Delta MNR$$

$$PM = MN$$ [By Construction]
$$QM = MR$$ [PM is the median]

$$\angle PMQ = \angle NMR$$ [Vertically opposite angles]
Therefore,
$$\Delta PQM = \Delta MNR$$ By SAS criterion of congruence
$$\Rightarrow PQ = RN$$ [CPCT] ... (ii)

Now,

$\begin{array}{l} \frac{{AB}}{{PQ}} = \frac{{AC}}{{PR}} = \frac{{AD}}{{PM}} & \left[ {{\rm{Given}}} \right]\\ \Rightarrow \frac{{CE}}{{RN}} = \frac{{AC}}{{PR}} = \frac{{AD}}{{PM}} & \begin{bmatrix} \text{from (i)} \\ \text{ and (ii)} \end{bmatrix} \\ \Rightarrow \frac{{CE}}{{RN}} = \frac{{AC}}{{PR}} = \frac{{2AD}}{{2PM}}\\ \Rightarrow \frac{{CE}}{{RN}} = \frac{{AC}}{{PR}} =\frac{{AE}}{{PN}} & \!\!\!\!\!\!\!\! \begin{bmatrix} 2AD = AE\\ {\text{ and }} \\ 2PM = PN \end{bmatrix} \\ \therefore \Delta ACE\sim\Delta PRN & \!\!\!\!\!\! \!\!\begin{bmatrix} \text{By SSS} \\ \text{ similarity} \\ \text{ criterion} \end{bmatrix} \end{array}$

Therefore,

\begin{align} \angle CAE &= \angle RPN\\{\rm{Similarly, }} \\ \angle BAE &= \angle QPN\\ \text{Therefore,} \\ \angle CAE + \angle BAE &= \begin{Bmatrix}\angle RPN \\+ \angle QPN \end{Bmatrix} \\ \Rightarrow \angle BAC &= \angle QPR\\ \Rightarrow \angle A &= \angle P....{\rm{(iii)}}\end{align}

\begin{align}&{\text{Now, In }}\Delta ABC{\text{ and }}\Delta PQR\\ &\frac{{AB}}{{PQ}} = \frac{{AC}}{{PR}}\\&\angle A = \angle P\qquad \left[ {{\rm{from (iii)}}} \right]\\&\therefore \Delta ABC\sim\Delta PQR \\ & \qquad \qquad \quad \begin{bmatrix} \text{By SAS} \\ \text{ similarity} \\ \text{ criterion} \end{bmatrix} \end{align}

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