in triangle ABC a-b-c and a-q-c prove the following
Answers
Answer:
We are given that in triangle ABC, segment PQ is parallel to side BC and it diveides the trangle ABC in two parts having area equal.
Therefore, Area triangle APQ= area quadrilateral PQCB
⇒Area triangle ABC= Area traingle APQ+ area quadrilateral PQCB
⇒Area triangle ABC=2×Area triangle APQ
⇒\frac{Area triangle ABC}{Area triangle APQ}=\frac{1}{2}
AreatriangleAPQ
AreatriangleABC
=
2
1
Now, by the property of similarity of triangles, we have
\frac{Area triangle APQ}{Area triangle ABC}=\frac{AP^{2} }{AB^{2}}=\frac{1}{2}
AreatriangleABC
AreatriangleAPQ
=
AB
2
AP
2
=
2
1
⇒\frac{AP}{AB}=\frac{1}{\sqrt{2} }
AB
AP
=
2
1
Now, we know that AP=AB-BP, therefore,
⇒\frac{AB-BP}{AB}=\frac{1}{\sqrt{2}}
AB
AB−BP
=
2
1
⇒1-\frac{BP}{AB}=\frac{1}{\sqrt{2} }1−
AB
BP
=
2
1
⇒1-\frac{1}{\sqrt{2}}=\frac{BP}{AB}1−
2
1
=
AB
BP
⇒\frac{BP}{AB}=\frac{\sqrt{2}-1}{\sqrt{2} }
AB
BP
=
2
2
−1
Answer:
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