Triangles | Exercise 6.3

Question 16

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

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

Solution

We will use the properties of similar triangles to prove the relationship between the medians.

Step 1 — List what we know

We are given that $\Delta$ ABC is similar to $\Delta$ PQR. This means their corresponding sides are in proportion.

$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}} \quad \text{... (1)}$

Also, their corresponding angles are equal.

$\angle \text{A} = \angle \text{P}$ $\angle \text{B} = \angle \text{Q} \quad \text{... (2)}$ $\angle \text{C} = \angle \text{R}$

We are told that AD is a median of $\Delta$ ABC. This means D is the midpoint of BC.

$\text{BD} = \frac{\text{BC}}{2} \quad \text{... (3)}$

We are also told that PM is a median of $\Delta$ PQR. This means M is the midpoint of QR.

$\text{QM} = \frac{\text{QR}}{2} \quad \text{... (4)}$

Diagram 1

Step 2 — Find a relationship for the bases

Let's use equations (1), (3), and (4). From equation (1), we know:

$\frac{\text{BC}}{\text{QR}} = \frac{\text{AB}}{\text{PQ}}$

Now, let's substitute $\text{BC} = 2 \times \text{BD}$ from (3). Let's also substitute $\text{QR} = 2 \times \text{QM}$ from (4).

$\frac{2 \times \text{BD}}{2 \times \text{QM}} = \frac{\text{AB}}{\text{PQ}}$

The 2s cancel out from the numerator and denominator.

$\frac{\text{BD}}{\text{QM}} = \frac{\text{AB}}{\text{PQ}} \quad \text{... (5)}$

Step 3 — Prove similarity of smaller triangles

Now, let's look at $\Delta$ ABD and $\Delta$ PQM. From equation (2), we know that:

$\angle \text{B} = \angle \text{Q}$

From equation (5), we just found that:

$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BD}}{\text{QM}}$

We have one equal angle and the sides including that angle are proportional. So, by the SAS similarity criterion:

$\Delta \text{ABD} \sim \Delta \text{PQM}$

Step 4 — Conclude the proof

Since $\Delta$ ABD is similar to $\Delta$ PQM, their corresponding sides are proportional. Therefore, we can write:

$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BD}}{\text{QM}} = \frac{\text{AD}}{\text{PM}}$

We have successfully shown the required relationship.

$\boxed{\frac{\text{AB}}{\text{PQ}} = \frac{\text{AD}}{\text{PM}}}$

Answer

The proof shows that the ratio of corresponding sides is equal to the ratio of corresponding medians.

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

In Fig. 6.36, $\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}}$ 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) $\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}$

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) $\frac{\text{CD}}{\text{GH}} = \frac{\text{AC}}{\text{FG}}$ (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 $\angle$ ADC = $\angle$ BAC. Show that $\text{CA}^2 = \text{CB.CD}$ .

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 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 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

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