Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.