Q) Triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer.
Here the largest triangle is represented by △ ABC and smallest one with shelf is represented by △ DEF. PQ is parallel to EF.
(i) Show that △ DPQ ∼ △ DEF.
(ii) If DP= 50 cm and PE = 70 cm then find PQ/EF.
(iii)(A) If 2 AB = 5DE and △ ABC ~ △ DEF then show that perimeter of △ ABC / perimeter of △ DEF is constant.
OR
(iii) (B) If AM and DN are medians of triangles ABC and DEF respectively then prove that △ ABM ∼ △ DEN.

Ans: (i) prove that ∠ DPQ ~ ∠ DEF:

Let’s compare the given triangles △ DPQ and △ DEF. Here we have:

Since we are given that PQ is parallel to EF, therefore:

∠ DQP = ∠ DEF      (corresponding angles)

∠ DPQ = ∠ DFE      (corresponding angles)

∠ PDQ = ∠ EDF      (common angles)

∴ by AAA similarity criterion, we get:

∠ DPQ ~ ∠ DEF ……… Hence Proved!

(ii) Value of :

We just proved, that ∠ DQP ~ ∠ DEF

Therefore,

∵ DE = DP + PE and given that DP = 50 cm, and PE = 70 cm,

∴ DE = 50 + 70 = 120 cm

∴ 

∴ 

(iii) (A) Perimeter of △ ABC / perimeter of △ DEF is constant. 

Since, it is given that △ ABC ~ △ DEF

Therefore, )

Since it is given that 2 AB = 5DE

∴

∴ 

∴ AB = DE; AC = x DF ; BC = x EF

Next,   =  

=

=

=

Therefore, perimeter of △ ABC / perimeter of △ DEF is constant. 

(iii) (B) prove that △ ABM ∼ △ DEN:

Let’s draw a diagram for the given condition:

Here AM is the median and bisects BC, ∴ BM = MC or BC = 2 BM

Similarly, DN is the median and bisects EF, ∴ EN = NF or EF = 2 EN

Next, we just proved that △ ABC ∼ △ DEF

∴

∴

∴

∠ B = ∠ E      (since △ ABC ∼ △ DEF)

∴ by SAS similarity criterion,

△ ABM ∼ △ DEN ……… Hence Proved!

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