Question

p and q are the midpoints of sides AC and a b respectively of triangle ABC a line through a is drawn parallel to BC meeting BP produced in line l and C Q produced in improve that area of triangle APL ISKCON is equal to area of triangle a qm second area of triangle A B L is equals to area of triangle AMC

Answer 1

Area of Δ APL = Area of Δ AQM

Step-by-step explanation:

in Δ APL & ΔCPB

AP = CP  ( P mid point of AC)

∠APL = ∠CPB  ( opposite angle)

∠ALP = ∠CPB  ( as AL ║ BC)

=>  Δ APL ≅ ΔCPB

=> Area of Δ APL = Area ofΔCPB

Similalrly

Δ AQM ≅ ΔBQC

Area of Δ AQM = Area of ΔBQC

Area of ΔCPB = (1/2) Area of ΔABC

Area of ΔBQC = (1/2) Area of ΔABC

=> Area of ΔCPB = Area of ΔBQC

=> Area of Δ APL = Area of Δ AQM

QED

Proved

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