Concurrency

Geometry Level pending

In △ A B C \triangle ABC , A B = 2 AB=2 and A C = 3 AC=3 .

Also the perpendicular bisector of side A B AB , the internal angle bisector of ∠ B A C \angle BAC and the side B C BC are concurrent.

Find the area of △ A B C \triangle ABC .


The answer is 2.958.

Pratik Shastri by Pratik Shastri , 35, India

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1 solution

Ujjwal Rane
Dec 11, 2014

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AB:AC = 2:3 and AP is the angle bisector hence AP:PC = 2:3. Take them to be 2x and 3x respectively.

PM perpendicular bisector of AB → P M = 4 x 2 − 1 \rightarrow PM = \sqrt{4x^2-1}

Drop PQ perpendicular to AC.

Triangles APM and BPM are congruent giving BP = AP = 2x and AQ = 1; QC = 2

P C = 4 x 2 + 3 PC = \sqrt{4x^2+3} which was originally taken as 3x

Equate and solve for x getting x = 3 5 x = \sqrt{\frac{3}{5}}

Area of ABC would be ( 2 + 3 ) 2 4 3 5 − 1 = 35 2 = 2.958 \frac{(2+3)}{2} \sqrt{4 \frac{3}{5}-1} = \frac{\sqrt{35}}{2} = \boxed{2.958}

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