Math Olympiad Test: Triangles - 3 - Class 10 MCQ

Let the length of sides of rhombus be x, length of OC be x1 and length of OD be y₁
Then, AB² + BC² + CD² + DA² = 4x²
Since, AC and DB bisect each other at O.
∴ AC = 2x₁ and BD = 2y₁
In ΔAOD, ΔDOC, ΔAOB, ΔBOC
4[x²₁ + y²₁] = 4x²
⇒ AC² + BD² = AB² + BC² + CD² + AD²

Since Δ’s ABC and DCB are similar,
∴ AB/AC = DC/DB ⇒ AB × DB = DC × AC

Since, 
∴ PQ || BC, [By converse of Thale’s theorem]
⇒ ∠APQ = ∠ABC and ∠AQP = ∠ACB (Corresponding angles)
∴ ΔAPQ ~ ΔABC [By AA similarity]

In Δ’s ABC and DAC,
∠BAC = ∠ADC (Given)
∠ACB = ∠ACD (Common angle)
∴ ΔABC ~ ΔDAC (By AA similarity)
⇒ AC/BC = DC/AC

Area of ΔABC = 1/2 × c × p = 1/2 × b × a

But, c² = a² + b² [By Pythagoras theorem]
∴ 

In ΔODB and ΔODC, using Pythagoras theorem,
OB² = OD² + BD² and OC² = OD² + CD²
∴ OC² = BD² – CD² ...(i)
Similarly, we have
OC² – OA² = CE² – AE² ...(ii)
And OA² – OB² = AF² – BF² ...(iii)
Adding (i), (ii) and (iii), we get
BD² + CE² + AF² – CD² – AE² – BF² = 0
⇒ BD² + CE² + AF² = CD² + AE² + BF²

Since, ΔDEF ~ ΔABC [Given]

∴ Perimeter of ΔDEF = DE + EF + FD
= 6 cm + 4 cm + 5 cm = 15 cm

In Δ's CDA and BAD,
we have
∠BAD = 90° – ∠ABD
∠DAC = 90° – ∠BAD
= 90° – (90° – ∠ABD) = ∠ABD
⇒ ∠DAC = ∠ABD

Now, in Δ's BDA and ADC, we have
∠ABD = ∠CAD
∠BDA = ∠ADC [Each 90°]
∴ ΔBDA ~ ΔADC [By AA similarity]
⇒ 

In ΔAOB and ΔCOD, we have
∠AOB = ∠COD
[Vertically opposite angles]

∠OAB = ∠OCD [Alternate interior angles]
∴ ΔAOB ~ ΔCOD [By AA similarity]
 [∵ AB = 2CD]

In ΔDBA and ΔDCB, we have
∠BDA = ∠CDB [Each 90°]
and ∠DBA = ∠DCB [Each = 90° – ∠A]
∴ ΔDBA ~ ΔDCB [By AA similarity]