NCERT Example 8 — Triangles

NCERT Example

Example 8

similar triangles · medians · SAS similarity · SSS similarity

CM and RN are respectively the medians of △ABC and △PQR. If △ABC ∼ △PQR, prove that:

(i) △AMC ∼ △PNR

(ii) CM/RN = AB/PQ

(iii) △CMB ∼ △RNQ

Given:

Triangle ABC is similar to triangle PQR; CM and RN are the corresponding medians.

To prove:

Parts (i) and (iii): the stated triangle similarities; part (ii): CM/RN = AB/PQ.

Proof:

Given △ABC ∼ △PQR, the correspondence is A ↔ P, B ↔ Q and C ↔ R.

Therefore, corresponding sides and angles are equal in ratio:

\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}\qquad ...(1)

\angle A=\angle P,\quad \angle B=\angle Q,\quad \angle C=\angle R

(i) Since CM and RN are medians:

AB=2AM,\qquad PQ=2PN

\frac{AM}{PN}=\frac{AB}{PQ}=\frac{CA}{RP}

Also, ∠MAC = ∠NPR because ∠A = ∠P. Therefore, △AMC ∼ △PNR by SAS similarity.

(ii) From △AMC ∼ △PNR:

\frac{CM}{RN}=\frac{CA}{RP}

From (1):

\frac{CA}{RP}=\frac{AB}{PQ}

Therefore:

\frac{CM}{RN}=\frac{AB}{PQ}

(iii) From (1) and part (ii):

\frac{CM}{RN}=\frac{BC}{QR}

Also:

BM=\frac{AB}{2},\qquad QN=\frac{PQ}{2}

\frac{BM}{QN}=\frac{AB}{PQ}=\frac{CM}{RN}

Thus:

\frac{CM}{RN}=\frac{BC}{QR}=\frac{BM}{QN}

Therefore, △CMB ∼ △RNQ by SSS similarity.

(i) △AMC ∼ △PNR; (ii) CM/RN = AB/PQ; (iii) △CMB ∼ △RNQ.