Question

In the triangle ABC IA and IB are bisectors of angles CAB and CBA respectively CP is parallel to IA and CQ is parallel to IB prove that: PQ is the perimeter of the triangle ABC

Answer 1

Answer:

Given data:

IA bisects ∠CAB  

IB bisects ∠CBA

CP // IA and CQ // IB

To prove: PQ = perimeter of triangle ABC

Step 1:

∠CAI = ∠IAB …… (i) [IA is given as an angle bisector]

Also, CP // IA, we have

∠IAB = ∠CPA …… (ii) [corresponding angles]

According to exterior angle property, we have

∠CAB = ∠CPA + ∠ACP

⇒ ∠CAI + ∠IAB = ∠CPA + ∠ACP

⇒ ∠CAI = ∠ACP ……. (iii)  [∵ ∠IAB = ∠CPA from (ii), therefore cancelling the similar terms]

From (i), (ii) & (iii), we get

∠CAI = ∠IAB = ∠CPA = ∠ACP …… (iv)

Similarly, we can also say that,

∠CBI = ∠IBA = ∠BCQ = ∠BQC …… (v)

Step 2 :

Let’s consider ∆ACP,  

∵ ∠CPA = ∠ACP  .... [from (iv)]

∴ AP = AC …… (vi)  [∵ sides opposite to equal angles are also equal]

Also, Consider ∆BCQ,

∵ angle BCQ = angle BQC  ...... [from (v)]

∴ BQ = BC …… (vii)  [∵ sides opposite to equal angles are also equal]

Step 3:

Now,  

PQ = AP + AB + BQ  

Or, PQ = AC + AB + BC …… [from (vi) & (vii)]

Since perimeter of a triangle the sum of all 3 sides of the triangle

∴ PQ = AB + BC + AC = Perimeter of ∆ ABC  

Hence proved