Question

In the given figure, l || m

(i) Name three pairs of similar triangles with proper correspondence; write similarities.
(ii) Prove that

\[\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{RQ}\]

Answer 1

Three pair of similar triangles are-

\[∆ ABK~ ∆ PQK\]   (AAA Similarity)

\[∆ CBK~ ∆ RQK\]  (AAA Similarity)

\[∆ ACK~ ∆ PRK\] (AAA Similarity)

(ii) Since the pair of similar triangles mentioned above can give us the desired result. The ratios of the corresponding side of the similar triangle are equal.

So,

`Δ ABK ≈ Δ PQK`

Therefore,

`(AB)/(PQ)=(AK)/(PK)=(BK)/(QK)` ……equation (1)

Similarly in  ,`Δ CBK ≈ Δ RQK` ……equation (2)

Similarly ,`Δ ACK ≈ Δ PRK`

`(AC)/(PQ)=(AK)/(PK)=(CK)/(RK)`……equation (3)

From the above equations 1 and 2 we have,

`(AB)/(PQ)=(AK)/(PK)=(BK)/(QK)=(CB)/(RQ)=(CK)/(RK)`

`⇒ (AB)/(PQ)=(CB)/(RQ)……eqaution (4)`

`⇒ (AB)/(CB)=(PQ)/(RQ)`

`⇒ (AB)/(CB)+1=(PQ)/(RQ)+1`

`⇒ (AC)/(CB)=(PR)/(RQ)`

`⇒ (AC)/(CB)=(PR)/(RQ)`

`⇒ (AC)/(PR)=(CB)/(RQ)`

Combining it with equation (4)

`(AC)/(PR)=(CB)/(RQ)=(AB)/(PQ)`

hence proved

`(AC)/(PR)=(AB)/(PQ)=(BC)/(RQ)`