Question

ABCD is a parallelogram. P is the midpoint of
side CD, seg BP meets diagonal AC at X. Prove that
Р
3AX = 2AC.​

Answer 1

In triangle ABX and triangle CPX angle ABX=angle CPX.......(alternate angles)

angle BAX=angle PCX.....(alternate angles)

Therefore triangle ABX is similar to triangle CPX by AA test of similarity

AB/CP=AX/CX.......corresponding sides of similar triangle

AB=2PC.....(AB=CD...opposite side if parallelogram and P is the midpoint of side CD)

AX+XC=AC XC=AC-AX

2PC/PC=AX/AC-AX.....(substitute the values)

2AC-2AX=AX

2AC=2AX+AX

2AC=3AX

Hence proved