Step-by-step explanation:
Proof :-
Given that
In ∆ ABC, PQ is a line drawn on AB and AC
In ∆ APQ and ∆ ABC
∠ APQ = ∠ ABC = 60°
∠ PAQ = ∠ BAC (Common vertex angle )
By AA similarity criteria
∆ APQ and ∆ ABC are similar triangles
=> ∆ APQ ~ ∆ ABC
We Know that
If two polygons of same number of sides are said to be similar,
If the corresponding angles are equal.
If the corresponding sides are in the same ratio or in proportion.
So, The corresponding angles are mist be in proportion
=> AP / PB = AQ/QC
Hence, Proved.
Used formulae:-
If two polygons of same number of sides are said to be similar,
If the corresponding angles are equal.
If the corresponding sides are in the same ratio or in proportion.
Note :-
If PQ || BC then by Basic Proportionality Theorem, AP/PB = AQ/QB
If a line drawn parallel to one side of a triangle intersecting other two sides at two different points and the line divides the other sides in the same ratio.
