Sides and and median of a triangle are respectively proportional to sides and and median of another triangle . Show that .

Understand the Question We are given two triangles and with medians and such that: To prove using the SAS similarity criterion, we have the ratio of two sides and need to establish that the included angles are equal, i.e., . Strategy: Extend medians and to form parallelograms and . Use SSS similarity on the sub-triangles ( and ) to show that and , which add up to give . Step 1 · Construct Parallelograms Extend median to point such that , and join and . Extend median to point such that , and join and . In quadrilateral , diagonals and bisect each other at . Therefore, is a parallelogram: Similarly, diagonals and bisect each other at , so is a parallelogram: Step 2 · Prove Similarity of Sub-Triangles Given Substitute , , , and By SSS similarity criterion Corresponding angles of similar triangles are equal Similarly, by proving Step 3 · Prove Adding equations and Now, in and By SAS similarity criterion Answer > Common Mistakes Confusing with Q12 (Given side instead of ): In Q12, the base side is given proportional to , allowing direct proof using . Here, is given instead of , which strictly requires constructing parallelograms. Assuming Medians are Altitudes: and are medians (bisecting the opposite sides), not altitudes or angle bisectors; do not assume or .

Triangles | Exercise 6.3

Question 14

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 $\Delta ABC \sim \Delta PQR$ .

Solution

Understand the Question

We are given two triangles $\Delta ABC$ and $\Delta PQR$ with medians $AD$ and $PM$ such that: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$
To prove $\Delta ABC \sim \Delta PQR$ using the SAS similarity criterion, we have the ratio of two sides $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ and need to establish that the included angles are equal, i.e., $\angle BAC = \angle QPR$ .
Strategy: Extend medians $AD$ and $PM$ to form parallelograms $ABEC$ and $PQLR$ . Use SSS similarity on the sub-triangles ( $\Delta ABE \sim \Delta PQL$ and $\Delta ACE \sim \Delta PRL$ ) to show that $\angle BAE = \angle QPL$ and $\angle CAE = \angle RPL$ , which add up to give $\angle BAC = \angle QPR$ .

Step 1 · Construct Parallelograms

Extend median $AD$ to point $E$ such that $AD = DE$ , and join $BE$ and $CE$ . Extend median $PM$ to point $L$ such that $PM = ML$ , and join $QL$ and $RL$ . Diagram 1

In quadrilateral $ABEC$ , diagonals $AE$ and $BC$ bisect each other at $D$ . Therefore, $ABEC$ is a parallelogram: $AC = BE \quad \text{and} \quad AB = CE$

Similarly, diagonals $PL$ and $QR$ bisect each other at $M$ , so $PQLR$ is a parallelogram: $PR = QL \quad \text{and} \quad PQ = LR$

Step 2 · Prove Similarity of Sub-Triangles

Given $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$

Substitute $AC = BE$ , $PR = QL$ , $AE = 2AD$ , and $PL = 2PM$

\begin{aligned} \dfrac{AB}{PQ} &= \dfrac{BE}{QL} = \dfrac{2AD}{2PM} \\[0.6em] \dfrac{AB}{PQ} &= \dfrac{BE}{QL} = \dfrac{AE}{PL} \end{aligned}

By SSS similarity criterion $\Delta ABE \sim \Delta PQL$

Corresponding angles of similar triangles are equal $\angle BAE = \angle QPL \quad \dots (1)$

Similarly, by proving $\Delta ACE \sim \Delta PRL$ $\angle CAE = \angle RPL \quad \dots (2)$

Step 3 · Prove $\Delta ABC \sim \Delta PQR$

Adding equations $(1)$ and $(2)$

\begin{aligned} \angle BAE + \angle CAE &= \angle QPL + \angle RPL \\[0.6em] \angle CAB &= \angle QPR \end{aligned}

Now, in $\Delta ABC$ and $\Delta PQR$

\begin{aligned} \dfrac{AB}{PQ} &= \dfrac{AC}{PR} \quad (\text{Given}) \\[0.6em] \angle CAB &= \angle QPR \end{aligned}

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

Answer

$\Delta ABC \sim \Delta PQR$

Common Mistakes

Confusing with Q12 (Given side $BC$ instead of $AC$ ): In Q12, the base side $BC$ is given proportional to $QR$ , allowing direct proof using $\Delta ABD \sim \Delta PQM$ . Here, $AC$ is given instead of $BC$ , which strictly requires constructing parallelograms.
Assuming Medians are Altitudes: $AD$ and $PM$ are medians (bisecting the opposite sides), not altitudes or angle bisectors; do not assume $\angle ADB = 90^\circ$ or $\angle BAD = \angle CAD$ .

Question 14

More questions in Exercise 6.3