Pl answer question 6

Dear Student,



We are given that AM, BN & CQ are altitudes of equal length, of sides BC, AC & AB respectively i.e.,
AM = BN = CQ  
And,
AM ⊥ BC
BN ⊥ AC
CQ ⊥ AB

Step 1:
Let’s consider in ∆ AMC and ∆ BNC,  
∠ C = ∠C [common angles]
∠AMC = ∠BNC = 90° [∵ BN & AM are perpendiculars to AC & BC]
AM = BN [given]
∴ By AAS criterion, we get
∆ AMC ≅ ∆ BNC

∴ AC = BC [∵ Corresponding parts of congruent triangles] ……. (i)

Step 2:
Let’s consider in ∆ AQC and ∆ ABN,
∠A = ∠ A [common angles]
∠ANB = ∠AQC = 90° [∵ BN & CQ are perpendiculars to AC & AB]
BN = CQ [given]
∴ By AAS criterion, we get
∆ AQC ≅ ∆ ABN

∴ AC = AB [Corresponding parts of congruent triangles] ……. (ii)

Step 3:
From (i) & (ii), we get
AB = BC = AC
Since triangles with all its sides having equal length is said to be an equilateral triangle
∴ ∆ ABC is an equilateral triangle
Hence proved
​​​
​Regards