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Ratios of Areas

Ratios of Areas

Lesson · Intermediate

Geometry: Plane Geometry

Triangle ABC with point P on BC and segment AP

\frac{[APB]}{[APC]}=\frac{BP}{PC}

Triangle ABC with point P on BC and point L on AP

\frac{[ALB]}{[ALC]}=\frac{BP}{PC}

Area Formulas and Important Geometry Formulas

Triangles APB and APC with common altitude AD

\frac{[APB]}{[APC]} = \frac{\frac12 AD\cdot BP}{\frac12 AD\cdot PC} = \frac{BP}{PC}

Triangle ABC with L on AP

\frac{[ABP]}{[APC]}=\frac{BP}{PC}=\frac{[BLP]}{[CLP]}

\frac{[ABP]}{[BLP]}=\frac{[APC]}{[CLP]}

\frac{[ABP]}{[ABP]-[BLP]}=\frac{[APC]}{[APC]-[CLP]}

\frac{[ABP]}{[ALB]}=\frac{[APC]}{[ALC]}

\frac{[ALB]}{[ALC]}=\frac{[ABP]}{[APC]}=\frac{BP}{PC}

In triangle ABC, D and E lie on BC and AC, respectively, such that

BD:DC=2:1

and

AE:EC=4:3

Let P be the intersection of AD and BE. If the area of ABC is 100, find the area of ABP.

Triangle ABC with cevians AD and BE intersecting at P

\frac{[BPD]}{[PDC]}=\frac{BD}{DC}=2

\frac{[APE]}{[PEC]}=\frac{AE}{EC}=\frac43

Triangle ABC with areas represented using w and t

\frac{[APB]}{[APC]}=\frac21

[APC]=4w+3w=7w

[APB]=2[APC]=14w

Triangle ABC with area ABP equal to 14w

\frac{[ABP]}{[BPC]}=\frac{AE}{EC}=\frac43

\frac{14w}{3t}=\frac43

\frac{7w}{t}=2

t=\frac{7w}{2}

Triangle ABC with all component areas determined

\begin{aligned} [ABC] &=14w+4w+3w+2t+t\\ &=21w+3\left(\frac{7w}{2}\right)\\ &=\frac{63w}{2} \end{aligned}

\frac{63w}{2}=100

w=\frac{200}{63}

[ABP]=14w=\boxed{\frac{400}{9}}

AMC 8 2019 (Problem 24)

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