Question

P and Q are the points on the side BC of triangle ABC and AP=AQ. Prove that AC+AB+BC>2AP+PQ

Answer 1

Step-by-step explanation:

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Answer 2

Concept:

If two triangles can be superimposed on one another, have equal-length sides, and equal-sized angles, they are said to be congruent. The following are the triangle congruence theorems or criteria that aid in demonstrating triangle congruence are SSS, SAS, ASA, AAS and RHS.

Given:

P and Q are the points on the side BC of triangle ABC and AP=AQ.

Find:

We have to prove that AC+AB+BC>2AP+PQ.

Solution:

In $\triangle$ ACQ, and $\triangle$ ABP, we have:

AP = AQ (already given)

AC = AB (by side)

CQ = PB (on the same base i.e. BC)

So, AC + CQ > AQ ..(1) (Sum of two sides of a triangle is greater than one side)

Similarly, AB + BP > AP .. (2) (Sum of two sides of a triangle is greater than one side)

On adding (1) and (2), we have:

AC + CQ + AB + BP > AQ + AP

AC + AB + (CQ + BP) > AQ + AP  

where, CQ + BP = BC, AQ = AP

AC + AB + BC > AP + AP

AC + AB + BC > 2AP + PQ (By proportionality theorem).

Hence, we have proved AC + AB + BC > 2AP + PQ.

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