In equilateral ∆abc with side a, the perpendicular to side ab at point b intersects extension of median am in point p. what is the perimeter of ∆abp, if mp = b?

Perimeter  = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

Now in ΔAMB , By using Pythagoras theorem

Now, in ΔBMP :

Perimeter of ABM = AB + PB + PM + AM

Hence, Perimeter of ΔABP = (2 + √3)·a units

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step-by-step explanation: