Triangles | Exercise 6.3

Question 12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\Delta PQR$ (see Fig. 6.41). Show that $\Delta ABC \sim \Delta PQR$ .

Solution

Understand the Question

We are given two triangles, $\Delta ABC$ and $\Delta PQR$ , where the sides $AB$ , $BC$ , and median $AD$ are proportional to $PQ$ , $QR$ , and median $PM$ respectively: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$
A median bisects the opposite side, so $BC = 2BD$ and $QR = 2QM$ .
Approach:
1. Replace $BC$ and $QR$ with $2BD$ and $2QM$ to show $\Delta ABD \sim \Delta PQM$ using SSS similarity.
2. Use this similarity to establish that $\angle B = \angle Q$ .
3. Finally, use the given ratio $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and $\angle B = \angle Q$ to prove $\Delta ABC \sim \Delta PQR$ by SAS similarity.

Step 1 · Relate Medians and Side Ratios

Given $AD$ is the median to side $BC$ and $PM$ is the median to side $QR$ . Diagram 1

Therefore, $D$ and $M$ are the midpoints of $BC$ and $QR$ respectively: $BD = \dfrac{1}{2}BC \implies BC = 2BD$ $QM = \dfrac{1}{2}QR \implies QR = 2QM$

Given $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$

Substituting $BC = 2BD$ and $QR = 2QM$ $\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AD}{PM}$

$\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$

Step 2 · Prove Similarity of $\Delta ABD$ and $\Delta PQM$

In $\Delta ABD$ and $\Delta PQM$ $\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$

By SSS similarity criterion $\Delta ABD \sim \Delta PQM$

Since corresponding angles of similar triangles are equal $\angle ABD = \angle PQM \implies \angle ABC = \angle PQR \quad \dots (i)$

Step 3 · Prove Similarity of $\Delta ABC$ and $\Delta PQR$

In $\Delta ABC$ and $\Delta PQR$ $\dfrac{AB}{PQ} = \dfrac{BC}{QR} \quad (\text{Given})$ $\angle ABC = \angle PQR \quad (\text{From equ. } (i))$

By SAS similarity criterion $\Delta ABC \sim \Delta PQR$

Answer

Hence proved, $\Delta ABC \sim \Delta PQR$ .

Common Mistakes

Assuming SAS Directly: Trying to apply SAS similarity directly to $\Delta ABC$ and $\Delta PQR$ without first proving $\angle B = \angle Q$ from the smaller triangles.
Median vs. Altitude: Confusing a median (which bisects the opposite side) with an altitude (which is perpendicular to the opposite side).
Incorrect Side Ratios: Replacing $BC$ with $BD$ without including the factor of $2$ ( $BC = 2BD$ ).

More questions in Exercise 6.3

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

In Fig. 6.35, $\Delta \text{ODC} \sim \Delta \text{OBA}$ , $\angle \text{BOC} = 125^\circ$ and $\angle \text{CDO} = 70^\circ$ . Find $\angle \text{DOC}$ , $\angle \text{DCO}$ and $\angle \text{OAB}$ .

Diagonals $\text{AC}$ and $\text{BD}$ of a trapezium $\text{ABCD}$ with $\text{AB} \parallel \text{DC}$ intersect each other at the point $\text{O}$ . Using a similarity criterion for two triangles, show that $\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}$ .

In Fig. 6.36, $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$ . Show that $\Delta \text{PQS} \sim \Delta \text{TQR}$ .

S and T are points on sides $PR$ and $QR$ of $\Delta \text{PQR}$ such that $\angle \text{P} = \angle \text{RTS}$ . Show that $\Delta \text{RPQ} \sim \Delta \text{RTS}$ .

In Fig. 6.37, if $\Delta \text{ABE} \cong \Delta \text{ACD}$ , show that $\Delta \text{ADE} \sim \Delta \text{ABC}$ .

In Fig. 6.38, altitudes $AD$ and $CE$ of $\Delta \text{ABC}$ intersect each other at the point $P$ . Show that:

(i) $\Delta \text{AEP} \sim \Delta \text{CDP}$

(ii) $\Delta \text{ABD} \sim \Delta \text{CBE}$

(iii) $\Delta \text{AEP} \sim \Delta \text{ADB}$

(iv) $\Delta \text{PDC} \sim \Delta \text{BEC}$

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta \text{ABE} \sim \Delta \text{CFB}$ .

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) $\Delta \text{ABC} \sim \Delta \text{AMP}$

(ii) $\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}$

Q10

CD and GH are respectively the bisectors of $\angle \text{ACB}$ and $\angle \text{EGF}$ such that D and H lie on sides AB and FE of $\Delta \text{ABC}$ and $\Delta \text{EFG}$ respectively. If $\Delta \text{ABC} \sim \Delta \text{FEG}$ , show that:

(i) $\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}$

(ii) $\Delta \text{DCB} \sim \Delta \text{HGE}$

(iii) $\Delta \text{DCA} \sim \Delta \text{HGF}$

Q11

In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$ . If $AD \perp BC$ and $EF \perp AC$ , prove that $\Delta ABD \sim \Delta ECF$ .

Q12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\Delta PQR$ (see Fig. 6.41). Show that $\Delta ABC \sim \Delta PQR$ .

Q13

D is a point on the side $BC$ of a triangle $ABC$ such that $\angle \text{ADC} = \angle \text{BAC}$ . Show that $\text{CA}^2 = \text{CB} \cdot \text{CD}$ .

Q14

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 $\Delta ABC \sim \Delta PQR$ .

Q15

A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.

Q16

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$ , respectively where $\Delta ABC \sim \Delta PQR$ , prove that

$\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}$